EMFT_Unit1_RGKNotes (1).pdf

Electromagnetic Field Theory (UE20EC303) A review of Vector Calculus Unit 1 Electrostatics Dr. Raghavendra G. Kulkarni Distinguis...

Electromagnetic Field Theory (UE20EC303) A review of Vector Calculus Unit 1 Electrostatics Dr. Raghavendra G. Kulkarni Distinguished Professor Department of Electronics and Communication Engineering PES University Bengaluru – 560085. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  z Position vectors and unit vectors in Cartesian coordinates r = x + y + z, z r r = x ax + y ay + z az , y y where x = x ax , y = y ay , z = z az , x (Vector quantities are in bold letters) x r, x, y, z are position vectors, which originate from origin; az ay ax, ay, az are unit vectors in x, y, z directions ax Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Distance vectors in Cartesian coordinates Let P (x1, y1, z1) and Q (x2, y2, z2) be two points in Cartesian z coordinate system. The position vectors of P and Q are RPQ Q rP = x1 ax + y1 ay + z1 az and rQ = x2 ax + y2 ay + z2 az P rP rQ and the distance vector starting from P to Q is RPQ y RPQ = rQ – rP = (x2 - x1)ax + (y2 - y1)ay + (z2 - z1)az x Example: Find distance vector, RPQ, if the points are P(1, 2 , 3) and Q(2, -2, 1) Solution: rP = 1 ax + 2 ay + 3 az , rQ = 2 ax - 2 ay + 1 az RPQ = rQ – rP = ax – 4 ay – 2 az Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Magnitude of a vector, unit vector Let a vector be: B = a x Bx + a y By + a z Bz The magnitude of B is: 𝑩 = 𝐵𝑥 2 + 𝐵𝑦 2 + 𝐵𝑧 2 The unit vector in the direction of B is: B / |B|, 𝑩 𝒂𝐵 = 𝐵𝑥 2 + 𝐵𝑦 2 + 𝐵𝑧 2 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Example: Specify the unit vector extending from the origin toward the point G (2, -2, -1). Solution: The vector extending from the origin to the point G, G = 2ax – 2ay – az The magnitude of G is, |G| = 4 + 4 + 1 = 3 The desired unit vector of G is, 𝑮 2 2 1 𝒂𝐺 = = 𝒂𝑥 − 𝒂𝑦 − 𝐚z |𝑮| 3 3 3 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  DOT product and its application A DOT product of two vectors: 𝑨 ∙ 𝑩 = 𝑨 𝑩 cos 𝜃𝐴𝐵 The angle, 𝜃𝐴𝐵 , is the smaller angle between A and B. Note that when 𝜃𝐴𝐵 is 90 ̊ the DOT product vanishes. θAB B If A = ax Ax + ay Ay + az Az and B = a x Bx + a y By + a z Bz 𝑨 ∙ 𝑩 = (ax Ax + ay Ay + az Az) ∙ (ax Bx + ay By + az Bz) B Since ax ∙ ax = 1, ax ∙ ay = 0, ax ∙ az = 0, and so on, 𝑨 ∙ 𝑩 = AxBx + AyBy + AzBz = A scalar quantity θBN aN Components of a vector in given direction: The magnitude of vector B (or the scalar component) in the direction of unit vector aN |B|cos θBN is given by the DOT product, B · aN = |B| cos θBN and the component vector of B in the direction of aN is: (B · aN) aN Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. The cross product of two vectors The cross product of two vectors, A and B, is defined as, 𝑨 × 𝑩 = 𝒂𝑁 𝑨 |𝑩| sin 𝜃𝐴𝐵 A 𝜃𝐴𝐵 where aN is the unit vector normal to the plane containing the B vectors A and B, in the direction of motion of right-hand screw from A to B. When 𝜃𝐴𝐵 is zero, the cross product vanishes. So, ax × ax = 0, ax × ay = az, ax × az = − ay, and so on. aN If A = ax Ax + ay Ay + az Az and B = ax Bx + ay By + az Bz , then the cross product can be determined as, 𝒂𝑥 𝒂𝑦 𝒂𝑧 𝑨 × 𝑩 = 𝐴𝑥 𝐴𝑦 𝐴𝑧 𝐵𝑥 𝐵𝑦 𝐵𝑧 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Examples using DOT product and cross product Example 1: Consider a vector, G = y ax – 2.5 x ay + 3 az , and the point Q(4, 5, 2). Find G at Q 1 The scalar component of G at Q in the direction of 𝒂𝑁 = (2𝒂𝑥 + 𝒂𝑦 − 2𝒂𝑧 ) 3 The vector component of G at Q in the direction of 𝒂𝑁 The angle between G at Q and 𝒂𝑁 Solution: G at Q is: G(Q) = 5 ax – 10 ay + 3 az , The scalar component of G at Q in the direction of 𝒂𝑁 is 1 1 G(Q) · 𝒂𝑁 = (5 ax – 10 ay + 3 az) ∙ (2𝒂𝑥 + 𝒂𝑦 − 2𝒂𝑧 ) = 10 − 10 − 6 = −2 3 3 The vector component of G at Q in the direction of 𝒂𝑁 −2 [G(Q) · 𝒂𝑁 ] 𝒂𝑁 = (2𝒂𝑥 + 𝒂𝑦 − 2𝒂𝑧 ) 3 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. The angle between G at Q and 𝒂𝑁 is obtained from G(Q) · 𝒂𝑁 = |G(Q)| cos 𝜃, G(Q) · 𝒂𝑁 cos 𝜃 = |G(Q)| −2 The magnitude of G at Q is: |G(Q)| = 25 + 100 + 9 = 134 , cos 𝜃 = 134 and hence 𝜃= 99.9.̊ Let us use cross product also to determine the angle. 1 G(Q) × 𝒂𝑁 = (17𝒂𝑥 + 16𝒂𝑦 + 25𝒂𝑧 ) 3 The magnitude is |G(Q) × 𝒂𝑁 |= 11.402, |G(Q)×𝒂𝑁 | sin 𝜃 = = 0.985 , |G(Q)||𝒂𝑁 | The angle 𝜃= 80.1 ̊, 99.9 ̊, so the correct angle is the common angle, 𝜃= 99.9 ̊. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Examples using DOT product and cross product Example 2: Find the cross product of two vectors, A = 2 ax – 3 ay + az and B = – 4 ax – 2 ay + 5 az and determine the angle between them. 𝒂𝑥 𝒂𝑦 𝒂𝑧 𝑨 × 𝑩 = 2 – 3 1 = −13𝑎𝑥 − 14𝑎𝑦 − 16𝑎𝑧 –4 –2 5 To find the angle between the vectors, the magnitudes of the two vectors, and the magnitude of their resultant cross product, have to be determined. The magnitude of A is 𝑨 = 4 + 9 + 1 = 14 and that of B is 𝑩 = 16 + 4 + 25 = 45. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. The magnitude of A X B is 𝑨 × 𝑩 = 169 + 196 + 256 = 621. Hence, 𝑨×𝑩 621 sin 𝜃𝐴𝐵 = = = 0.993, so, 𝜃𝐴𝐵 = 83.13 ̊, 96.87 ̊ 𝑨 𝑩 630 Note that two angles have emerged, and to determine the correct angle, we use the DOT product. 𝑨 ∙ 𝑩 = 2𝒂𝑥 − 3𝒂𝑦 + 𝒂𝒛 ∙ −4𝒂𝑥 − 2𝒂𝑦 + 5𝒂𝒛 = 3 𝑨∙𝑩 𝜃𝐴𝐵 = cos −1 = cos −1 0.1195 = 83.13° 𝑨 |𝑩| Hence the common angle is, 83.13 ,̊ which is the correct angle. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Circular cylindrical coordinate system In circular cylindrical coordinate system, the point P in space is defined by ρ, φ, and z. z az aφ ρ is the perpendicular distance of point P from Q ρ z-axis (QP); φ is the angle between x-axis and z the line segment OP’ formed by projection of P P (ρ, φ, z) on xy-plane; z is same as in Cartesian system. aρ Transformation from cylindrical coordinates to Cartesian coordinates, and vice versa: y y O From the figure note that x φ ρ x = ρ cos(φ) , y = ρ sin(φ) x P’ ρ = 𝑥 2 + 𝑦 2 , φ = 𝑡𝑎𝑛−1 (y/x) Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Circular cylindrical coordinate system The Unit vectors in cylindrical coordinate system are: z az aρ, aφ, az and they are mutually perpendicular, and aφ their cross products are as below. Q ρ z aρ x a φ = az , aφ x az = a ρ , az x aρ = aφ P (ρ, φ, z) aρ Note that aρ is in the direction of outward normal to the constant- ρ cylindrical surface (normal to z-axis, see figure); O y ρ aφ is normal to the constant-φ plane (OQPP’ in the φ figure), towards increasing φ angle. x P’ Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Circular cylindrical coordinate system aφ DOT products between unit vectors of cylindrical and Cartesian systems aρ · ax = | aρ || ax |cos(φ) = cos(φ) az ay aρ · ay = | aρ || ay |cos(90 – φ) = sin(φ) 90 - φ φ aρ · az = | aρ || az |cos(90) = 0 aρ ax aφ · ax = | aφ|| ax |cos(90 + φ) = – sin(φ) aφ · ay = | aφ|| ay |cos(φ) = cos(φ) Note that + z-axis is coming out from the aφ · az = | aφ|| az |cos(90) = 0 screen Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Circular cylindrical coordinate system Example: Transform the vector 𝑩 = 𝑦𝒂𝑥 − 𝑥𝒂𝑦 + 𝑧𝒂𝒛 into cylindrical coordinates. Solution: In cylindrical coordinates, 𝑩 = 𝐵𝜌𝒂𝜌 + 𝐵𝜑𝒂𝜑 + 𝐵𝑧𝒂𝑧 , where 𝐵𝜌 , 𝐵𝜑, and 𝐵𝑧 are scalar components of B in ρ, φ, and z directions, which are to be determined. 𝐵𝜌 = 𝑩 ∙ 𝒂𝜌 = 𝑦𝒂𝑥 − 𝑥𝒂𝑦 + 𝑧𝒂𝒛 ∙ 𝒂𝜌 = 𝑦 𝒂𝑥 ∙ 𝑎𝜌 − 𝑥 𝒂𝑦 ∙ 𝒂𝜌 = 𝑦 cos 𝜑 − 𝑥 sin 𝜑 = 𝜌 sin 𝜑 cos 𝜑 − 𝜌 cos 𝜑 sin 𝜑 = 0 𝐵𝜑 = 𝑩 ∙ 𝒂𝜑 = 𝑦𝒂𝑥 − 𝑥𝒂𝑦 + 𝑧𝒂𝒛 ∙ 𝒂𝜑 = 𝑦 𝒂𝑥 ∙ 𝒂𝜑 − 𝑥 𝒂𝑦 ∙ 𝒂𝜑 = −𝑦 sin 𝜑 − 𝑥 cos 𝜑 = −𝜌𝑠𝑖𝑛2 𝜑 − 𝜌𝑐𝑜𝑠 2 𝜑 = −𝜌 Hence B in cylindrical coordinates is: 𝑩 = −𝜌𝒂𝜑 + 𝑧𝒂𝑧 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system z z’ ar In spherical coordinates, the point P in space is defined by the coordinates, r, θ, aφ and φ. P (r, θ, φ) r The distance OP from origin is r. The angle made by the line OP with z-axis is θ. The θ θ aθ angle made by the projection of OP on xy- plane with x-axis is φ. O y Note that the unit vectors, ar a𝜃 and aφ φ are mutually perpendicular. x Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system The direction of ar is the direction of z z’ ar position vector r. The direction of a𝜃 is that of perpendicular to OP in the increasing side aφ of θ. The direction of aφ is that of normal to P (r, θ, φ) the plane formed by z-axis and OP. Notice r that both ar and a𝜃 are in this plane (and P’ they are mutually perpendicular). θ θ aθ In order to transform a vector given in spherical coordinates to vector in cartesian O φ y coordinates, one should know the DOT O’ products between unit vectors in one x system to that in the other system. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system DOT product, ar · ax : Consider the plane z formed by z-axis and the line P’O”. Unit Z’ ar vectors ar , a𝜃 are in this plane, while aφ is θ perpendicular to the plane. aφ P Notice that angle between ar and PZ’ is θ, so angle between ar and PP’ is 90 – θ (since PZ’ P’ θ θ and PP’ are perpendicular). Hence projection aθ of ar on PP’ is cos(90 – θ), which is sin(θ). O Since PP’ is parallel to OO”, projection of ar y on OO” is also sin(θ). As OO” makes an angle φ O’ φ with x-axis, further projection on x-axis will x O” be sin(θ) cos(φ), so, ar · ax = sin(θ) cos(φ) Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system DOT products, ar · ay ; ar · az z Z’ ar We know from the previous slide that projection of ar on OO” is sin(θ). As OO” aφ makes an angle 90 – φ with y-axis, further P projection on y-axis will be sin(θ) cos(90 – φ). Thus, P’ ar · ay = sin(θ) sin(φ) θ θ aθ Since angle between ar and az is θ, we get, O y ar · az = cos(θ) φ O’ x O” Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system DOT product, a𝜃 · ax z Z’ ar The angle between ar and a𝜃 (so OP and a𝜃 ) is 𝜃 90 ;̊ so the angle between a𝜃 and PO’ is 90 – θ. aφ Since PO’ and PP’ are perpendicular, the angle P between aθ and PP’ is θ. Hence projection of aθ on PP’ is cos(θ). 𝜃 P’ 𝜃 𝜃 aθ Since PP’ is parallel to OO”, projection of a𝜃 on OO” is also cos(θ). As OO” makes an angle φ O with x-axis, further projection on x-axis will be y cos(θ) cos(φ), Thus, φ O’ x O” a𝜃 · ax = cos(θ) cos(φ) Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system DOT products, a𝜃 · ay and a𝜃 · az z Z’ ar From the earlier discussion, we know that θ projection of a𝜃 on OO” is cos(θ). As OO” makes aφ an angle 90 – φ with y-axis, further projection on P y-axis will be cos(θ) cos(90 – φ), Thus, θ P’ a𝜃 · ay = cos(θ) sin(φ) θ θ aθ The angle between aθ and az is 90 + θ; hence O y a𝜃 · az = cos(90 + θ) = – sin(θ) φ O’ x O” Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system DOT products, a∅ · ax ; a∅ · ay ; a∅ · az z We know that a∅ is perpendicular to the plane Z’ ar formed by z-axis the line OP. This plane makes θ angle φ with x-axis. So the angle between a∅ and aφ ax is 90 + φ. Therefore, P a∅ · ax = cos(90 + φ) = – sin(φ) θ P’ θ θ aθ Since the angle between a∅ and ax is 90 + φ, the angle between a∅ and ay is φ. Therefore, O y a∅ · ay = cos(φ) φ O’ x O” Since a∅ is perpendicular to az the DOT product, a∅ · az = 0. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Spherical coordinate system Scalar transformation of coordinates: Note that OO’ = QP = r sin(θ), and z z’ ar x = OO’ cos(φ), hence x = r sin(θ) cos(φ) Q aφ Similarly, y = OO’ sin(φ), hence P y = r sin(θ) sin(φ), Note that z = OQ = r cos(θ). Using these r relations, the reverse transformations are, θ θ aθ 𝑟= 𝑥2 + 𝑦2 + 𝑧2 O 𝑧 φ y 𝜃= cos −1 O’ 𝑥 2 +𝑦 2 +𝑧 2 x 𝑦 𝜑 = tan−1 𝑥 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  𝑥𝑧 Example: Express the vector, 𝑮= 𝒂𝑥 in components of spherical 𝑦 coordinates. Solution: First we express 𝑮 = 𝐺𝑟 𝒂𝑟 + 𝐺𝜃 𝒂𝜃 + 𝐺𝜑 𝒂𝜑 , where 𝐺𝑟 = 𝑮 ∙ 𝒂𝑟 , 𝐺𝜃 = 𝑮 ∙ 𝒂𝜃 , 𝐺∅ = 𝑮 ∙ 𝒂∅ 𝑥𝑧 𝑥𝑧 𝐺𝑟 = 𝒂𝑥 ∙ 𝒂𝑟 = sin 𝜃 cos 𝜑 𝑦 𝑦 Using coordinate transformations, 𝑥 = 𝑟 sin 𝜃 cos ∅ , 𝑦 = 𝑟 sin 𝜃 sin ∅, 𝑧 = 𝑟 cos 𝜃, 𝑐𝑜𝑠 2 ∅ 𝐺𝑟 = 𝑟 sin 𝜃 cos 𝜃 sin ∅ 2 𝑥𝑧 𝑥𝑧 2 𝑐𝑜𝑠 ∅ 𝐺𝜃 = 𝒂𝑥 ∙ 𝒂𝜃 = cos 𝜃 cos ∅ = 𝑟 𝑐𝑜𝑠 𝜃 𝑦 𝑦 sin ∅ 𝑥𝑧 𝑥𝑧 𝐺∅ = 𝒂𝑥 ∙ 𝒂∅ = (− sin ∅) = −𝑟 cos 𝜃 cos ∅ 𝑦 𝑦 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Introducing the concept of the ‘Del’ operation Consider a scalar function, f (x, y, z), in space. Let an incremental change in f be df in an incremental length, dl, where the vector, 𝒅𝒍 = 𝒂𝑥 𝑑𝑥 + 𝒂𝑦 𝑑𝑦 + 𝒂𝑧 𝑑𝑧 ; note that df is defined as, 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝑑𝑓 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝑑𝑓 = 𝒂𝑥 + 𝒂𝑦 + 𝒂𝑧 ∙ 𝒂𝑥 𝑑𝑥 + 𝒂𝑦 𝑑𝑦 + 𝒂𝑧 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕 𝜕 𝜕 = 𝒂𝑥 + 𝒂𝑦 + 𝒂𝑧 𝑓 ∙ 𝒂𝑥 𝑑𝑥 + 𝒂𝑦 𝑑𝑦 + 𝒂𝑧 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 Let us denote 𝜕 𝜕 𝜕 𝜵 = 𝒂𝑥 + 𝒂𝑦 + 𝒂𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 which is known as Del operator. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Hence df becomes, 𝑑𝑓 = 𝜵𝑓 ∙ 𝒅𝒍 = 𝜵𝑓 𝒅𝒍 cos(𝜃) Notice that when dl aligns with 𝜵𝑓 (means θ is zero) the change df is maximum. Hence 𝜵𝑓 indicates the direction of maximum change in the scalar function, f(x, y, z). Gradient of scalar function, 𝜵𝑓 : 𝜵𝑓 is known as gradient of function f, or Grad (f), and its direction indicates the direction of maximum change in f. The direction normal to 𝜵𝑓 is the direction of zero change in f, or the contour of constant f. Thus 𝜵𝑓 = 0 yields the surface of constant f. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Divergence of vector function: The divergence of vector at a point, 𝜵 · 𝑫 , is the net outflow of that vector from an infinitesimal volume about that point. Divergence Theorem: න 𝜵 ∙ 𝑫 𝑑𝑣 = ර 𝑫 ∙ 𝒅𝒔 𝑉𝑜𝑙 𝑆 Curl of a vector: The curl of a vector, A, is given by 𝜵 x A and in cartesian coordinates it is, 𝒂𝑥 𝒂𝑦 𝒂𝑧 𝜕 𝜕 𝜕 𝜵×𝑨= 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝐴𝑥 𝐴𝑦 𝐴𝑧 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics The electric force exerted by charge Q1 on charge Q2 is: 𝑄1 𝑄2 𝑄1 𝑄2 𝑭12 = 2 𝒂12 = 3 𝑹12 , z 4𝜋∈ 𝑅12 4𝜋∈ 𝑅12 Q2 F12 Q1 R12 See the Figure on the left: r1 : position vector of Q1 point r2 r2 : position vector of Q2 point r1 R12 : distance vector from Q1 to Q2 y Є : permittivity of the medium. For free space, O 10−9 Є = Є0 = 𝐹/𝑚 36𝜋 x R12 = r2 – r1 ; a12 = R12 / R12 where a12 is unit vector in the direction of R12 and R12 is the magnitude of R12. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics The electric force exerted by the charges, Q1 , Q2 , Q3 ,... on the charge QA , is given by, 𝑄1 𝑄𝐴 𝑄2 𝑄𝐴 𝑄3 𝑄𝐴 𝑭𝐴 = 𝑭1𝐴 + 𝑭2𝐴 + 𝑭3𝐴 + ⋯ = 𝒂 + 𝒂 + 𝒂 +.... 4𝜋∈𝑅1𝐴 2 1𝐴 4𝜋∈𝑅2𝐴 2 𝟐𝐴 4𝜋∈𝑅3𝐴 2 𝟑𝐴 𝑄1 𝑄𝐴 where 𝑭1𝐴 = 𝒂 , 4𝜋∈𝑅1𝐴 2 1𝐴 R1A = rA – r1 , a1A = R1A / R1A , and so on. where a1A is unit vector in the direction of R1A and R1A is the magnitude of R1A. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics Example: Three point-charges are located at the corners of a rectangle in free space as shown in the figure. The charges are: Q1 = 3 µC, Q2 = – 2 µC, and Q3 = 5 µC. Find Q3 the net force on Q3. Q2 Solution: Let Q1 be at origin. So, r1 = 0, r2 = 0.03 ay 3 cm and r3 = 0.04 ax + 0.03 ay. Force exerted by Q1 on Q3 is Q1 𝑄1 𝑄3 y 4 cm 𝑭13 = 3 𝑹13 , where 4𝜋∈0 𝑅13 x R13 = r3 – r1 = 0.04 ax + 0.03 ay , and R13 = 0.05 m. Since for free space, Є0 = 10-9/36π, we obtain, F13 = 43.2 ax + 32.4 ay Newtons. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics Similarly, R23 = r3 – r2 = 0.04 ax , and R23 = 0.04 m. F13 Force exerted by Q2 on Q3 is Q2 F23 Q3 𝑄2 𝑄3 𝑭𝟐3 = 3 𝑹23 , 3 cm 4𝜋∈0 𝑅23 Thus, F23 = – 56.25 ax Newtons. Hence the net force Q1 4 cm on Q3 is: y F3 = F13 + F23 = – 13.05 ax + 32.4 ay Newtons. x Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics Electric field: The electric field intensity is the force exerted by the charge Q on a unit positive charge. z E 𝑄 𝑄(𝒓 − 𝒓′ ) P 𝑬= 2 𝒂𝑅 = R 4𝜋 ∈ |𝒓 − 𝒓′| 4𝜋 ∈ |𝒓 − 𝒓′|3 Q r r : position vector of the field point, P r’ r’ : position vector of the point, where charge Q is placed y R : distance vector from the charge Q to the field point, O and R = r – r’ aR : Unit vector in the direction of R x Є : permittivity of the medium If the charge is placed at origin, r’ = 0, and R = r, then E is: 𝑄 𝑄𝒓 𝑬= 2 𝒂𝑅 = 4𝜋 ∈ |𝒓| 4𝜋 ∈ |𝒓|3 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics If there are more point charges, the net electric field Q2 is the vector sum of electric fields due to individual z charges. P E1 Q1 𝑁 E2 1 𝑄𝑘 (𝒓 − 𝒓𝑘 ) r2 𝑬= ෍ r 4𝜋 ∈ |𝒓 − 𝒓𝑘 |3 r1 𝑘=1 y r : position vector of the field point O rk : position vector of the point, where charge Qk is x placed Є : permittivity of the medium E1 : Field due to charge Q1 E2 : Field due to charge Q2 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics If there is a charge density of ρL C / m over a length of a wire, then the charge in an infinitesimal length dl is: dQ = ρL dl , + + and total charge is: + + + + 𝑄 = න 𝜌𝐿 𝑑𝑙 + + If a charge density of ρs C / m2 is defined over a surface, then the charge in an infinitesimal surface area ds is: dQ = ρs ds , and total charge is: + + + + + ++ 𝑄 = ඵ ρs ds + + + If a charge density of ρv C / m3 is defined in a volume, then the charge in an infinitesimal volume dv is: dQ = ρv dv , and total + + + + + + + charge is: + + + 𝑄 = ම ρv dv Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics + + For a line charge, the electric field is: + + 𝑬= 1 න 𝜌𝐿 𝑹𝑑𝐿′ + + 4𝜋𝜖 𝑅3 + R = r - r’ where R = r – r’ is the distance vector from the + dL’ P elemental charge, ρL dL’, to the field point, and R is r’ r magnitude of R. R P + + ++ +dS’ + + In case of surface charge, the electric field is: + + + 𝑬= 1 න 𝜌𝑠 𝑹𝑑𝑆′ 4𝜋𝜖 𝑅3 P For a volume charge, the electric field is: + + + R + + + dv’ + 1 𝜌𝑣 𝑹𝑑𝑣′ ++ + 𝑬= 4𝜋𝜖 න 𝑅3 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Example: Find the total charge in the cylinder of height 30 cm and radius 10 cm, if the charge density inside the cylinder is given by: 𝜌𝑣 = 100𝑒 −𝑧 𝑥 2 + 𝑦 2 −0.25 C/m3 Solution: The problem can be solved easily in cylindrical z coordinates. Hence the charge density is expressed as: 𝜌𝑣 = 100𝑒 −𝑧 𝑥 2 + 𝑦 2 −0.25 = 100𝑒 −𝑧 𝜌−0.5 ρ since ρ2 = x2 + y2 , total charge enclosed in the cylinder is: z 𝑄 = ‫ 𝑣𝑑 𝑣𝜌 ׮‬, where 𝑑𝑣 = 𝜌 𝑑𝜌𝑑𝜑𝑑𝑧, the infinitesimal y volume in cylindrical coordinates. So, 𝑄 = ‫ = 𝑧𝑑𝜑𝑑𝜌𝑑 𝜌 𝑣𝜌 ׮‬100 ‫ 𝑒 ׮‬−𝑧 𝜌+0.5 𝑑𝜌𝑑𝜑𝑑𝑧 , φ x The limits of integration are: ρ = 0 to 0.1 m, φ = 0 to 2π rad, and z = 0 to 0.3 m. Integrating, we get Q = 3.43 C. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Example: A charge of Q is uniformly distributed over a thin wire of length L. Determine the field at the point P shown in the figure. y L a x P x dx Solution: Let ρL C / m be the uniform charge density on the wire. Hence charge in the infinitesimal length dx will be (ρL dx) C. Infinitesimal field produced by this charge is, 𝜌𝐿 𝑑𝑥 𝑹 𝒅𝑬 = 4𝜋𝜖 𝑅3 where R = r – r’ = (L + a) ax – x ax = (L + a – x) ax , and R = (L + a – x). Hence, 𝜌𝐿 𝑑𝑥 𝒅𝑬 = 2 𝐚𝑥 4𝜋𝜖 (𝐿 + 𝑎 − 𝑥) Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics The net electric field due to total charge on the wire can be obtained by integrating elemental field as below. 𝜌𝐿 𝐿 𝑑𝑥 𝜌𝐿 1 𝐿 𝑬 = 𝐚𝑥 න 2 = 𝐚𝑥 4𝜋𝜖 0 (𝐿 + 𝑎 − 𝑥) 4𝜋𝜖 𝐿 + 𝑎 − 𝑥 0 𝜌𝐿 1 1 𝜌𝐿 𝐿 𝑬 = 𝐚𝑥 − = 𝐚𝑥 4𝜋𝜖 𝑎 𝐿 + 𝑎 4𝜋𝜖 𝑎(𝐿 + 𝑎) Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics Example: A continuous volume charge, 𝜌𝑣 = (𝑥 2 + 𝑦 2 + 𝑧 2 )5Τ2 , is distributed in the region, 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , 0 ≤ z ≤ 1 , and is zero elsewhere. Find the electric field at the origin (see figure). Solution: Let an elemental volume dv be at point (x, y, z) z within the cube as shown. Notice that field point is at origin. So, r = 0, and r’ = x ax + y ay + z az , hence (x, y, z) R = r – r ‘ = – ( x ax + y ay + z az ), and R y 𝑅 = (𝑥 2 + 𝑦 2 + 𝑧 2 )1Τ2 x For a volume charge, the electric field is: 1 𝜌𝑣 𝑹𝑑𝑣 𝑬= න 4𝜋𝜖 𝑅3 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics 111 −1 𝑬= ම 𝑥2 + 𝑦2 + 𝑧2 𝑥𝒂𝑥 + 𝑦𝒂𝑦 + 𝑧𝒂𝑧 𝑑𝑥𝑑𝑦𝑑𝑧 4𝜋𝜖 0 0 0 Let us determine x component of electric field, Ex. 111 −1 𝐸𝑥 = ම [ 𝑥 3 + 𝑥𝑦 2 + 𝑥𝑧 2 𝑑𝑥] 𝑑𝑦𝑑𝑧 4𝜋𝜖 0 0 0 Integrating with respect to x first and applying limits, 11 −1 1 𝑦2 𝑧2 𝐸𝑥 = ඵ + + 𝑑𝑦 𝑑𝑧 4𝜋𝜖 4 2 2 00 Integrating with respect to y and then z (and applying limits), −1 1 5 𝑧 2 −1 7 𝐸𝑥 = න + 𝑑𝑧 = = −5.25 × 109 𝑉/𝑚 4𝜋𝜖 0 12 2 4𝜋𝜖 12 Due to symmetry, Ey and Ez components also have same magnitude as Ex. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Gauss Law: Gauss law states that the total flux, 𝜓, passing through any closed surface is equal to the total charge, 𝑄, enclosed by the surface. 𝜓 = 𝑄 = න 𝜌𝑣 𝑑𝑣 𝑉𝑜𝑙 If D is the flux density (or displacement density) at a point on a surface, then the flux passing through an elemental surface ds is, 𝑑𝜓 = 𝑫 ∙ 𝒅𝒔 Total flux passing through closed surface is then expressed as, 𝜓 = ර 𝑫 ∙ 𝒅𝒔 𝑆 ර 𝑫 ∙ 𝒅𝒔 = 𝑄 = න 𝜌𝑣 𝑑𝑣 𝑆 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Divergence Theorem: From divergence theorem, we know that ර 𝑫 ∙ 𝒅𝒔 = න 𝜵 ∙ 𝑫 𝑑𝑣 𝑺 𝑉𝑜𝑙 Therefore, න 𝜵 ∙ 𝑫 𝑑𝑣 = න 𝜌𝑣 𝑑𝑣 = 𝑄 𝑉𝑜𝑙 𝑉𝑜𝑙 Hence, 𝜵 ∙ 𝑫 = 𝜌𝑣 Divergence of displacement density at a point is equal to the volume charge density at that point. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Work: The force experienced by a charge Q in an electric field E is QE, and the work done against this force in moving the charge a small distance dl is: dW = – QE · dl Hence total work done in moving the charge from point B to point A against the field is: 𝐴 𝑊 = −𝑄 න 𝑬 ∙ 𝒅𝒍 𝐵 Potential difference: The work done against the electric field in moving a unit charge from point B to point A is termed as potential difference between A and B. 𝐴 𝑉𝐴𝐵 = 𝑊/𝑄 = − න 𝑬 ∙ 𝒅𝒍 𝐵 Traversing through any path, if final point A coincides with initial point B, then there is no potential difference. Hence, ර 𝑬 ∙ 𝒅𝒍 = 0 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Potential: The work done against the electric field in moving a unit charge from infinity to point A is termed as potential at point A. 𝐴 𝐵 𝑉𝐴 = − න 𝑬 ∙ 𝒅𝒍 , similarly 𝑉𝐵 = − න 𝑬 ∙ 𝒅𝒍 ∞ ∞ The potential difference between A and B is, by definition, 𝐴 𝑉𝐴𝐵 = − න 𝑬 ∙ 𝒅𝒍 𝐵 However, using the definition of potentials, it can be expressed as, 𝐴 𝐵 𝐵 𝐴 𝑉𝐴𝐵 = 𝑉𝐴 − 𝑉𝐵 = − න 𝑬 ∙ 𝒅𝒍 − − න 𝑬 ∙ 𝒅𝒍 = න 𝑬 ∙ 𝒅𝒍 − න 𝑬 ∙ 𝒅𝒍 ∞ ∞ ∞ ∞ Hence, 𝐵 𝐴 𝑉𝐴𝐵 = 𝑉𝐴 − 𝑉𝐵 = න 𝑬 ∙ 𝒅𝒍 = − න 𝑬 ∙ 𝒅𝒍 𝐴 𝐵 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Potential due to a point charge at origin: Consider a point charge Q placed at origin. The electric field at a distance r from the origin is z 𝑄 ar 𝑬= 𝒂𝑟 4𝜋𝜖𝑟 2 r Q y The incremental distance in radial direction is, dl = dr ar , x since there is no change in θ or φ, as the unit charge is moved from ∞ to point A in radial direction; hence the potential is, 𝐴 −𝑄 𝑟 1 𝑄 𝑉 = − න 𝑬 ∙ 𝒅𝒍 = න 2 𝑑𝑟 = ∞ 4𝜋𝜖 ∞ 𝑟 4𝜋𝜖𝑟 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics z Potential due to a point charge in space at r’: Q r - r’ P The potential at point P is, r’ r 𝑄 𝑉= 4𝜋𝜖|𝒓 − 𝒓′ | y Potential due to several point charges in space: x The potential at point P is, 𝑁 1 𝑄𝑘 𝑉= ෍ 4𝜋𝜖 |𝒓 − 𝒓𝑘 | 𝑘=1 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Electrostatics dv’ r - r’ Potential due to continuous volume charge distribution: r’ r 1 𝜌𝑣 𝑑𝑣′ y 𝑉= න 4𝜋𝜖 𝑉𝑜𝑙 |𝒓 − 𝒓′ | x dS’ r - r’ Potential due to continuous surface charge distribution: r’ r 1 𝜌𝑠 𝑑𝑆′ y 𝑉= න 4𝜋𝜖 𝑆 |𝒓 − 𝒓′ | x dL’ r - r’ Potential due to continuous line charge distribution: r’ r 1 𝜌𝐿 𝑑𝐿′ 𝑉= න y 4𝜋𝜖 𝐿 |𝒓 − 𝒓′ | x Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Electric field in terms of potential : The incremental work in moving a unit charge for an incremental distance is called incremental potential. 𝑑𝑉 = −𝑬 ∙ 𝒅𝒍 However, dV can be expressed as, 𝜕𝑉 𝜕𝑉 𝜕𝑉 𝑑𝑉 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕 𝜕 𝜕 = 𝒂𝑥 + 𝒂𝑦 + 𝒂𝑧 𝑉 ∙ 𝒂𝑥 𝑑𝑥 + 𝒂𝑦 𝑑𝑦 + 𝒂𝑧 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 Thus, 𝑑𝑉 = 𝜵𝑉 ∙ 𝒅𝒍 = −𝑬 ∙ 𝒅𝒍 which yields, 𝑬 = −𝜵𝑉 , performing curl operation, 𝛁 × 𝑬 = −𝛁 × 𝜵𝑉 = 0 Since Curl of a Gradient is zero (a vector identity) Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Maxwell’s equations for electrostatic case: 𝛁 × 𝑬 = 0, 𝛁 ∙ 𝑫 = 𝜌𝑣 Energy stored in the electric field: The energy stored in the electric field is: 1 1 𝑊𝐸 = න 𝜌𝑣 𝑉 𝑑𝑣 = න ∈0 𝐸 2 𝑑𝑣 2 𝑉𝑜𝑙 2 𝑉𝑜𝑙 From the above expression, we note the quantity inside the volume integral has the dimension of density, so we define an energy density as: 1 𝑤𝐸 = ∈0 𝐸 2 2 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Electrostatics Example: A potential field, 𝑉 = 2𝑥 2 𝑦 − 5𝑧, exists in free space. Determine potential, electric field, direction of electric field, displacement density, charge density, and energy density, at point P( – 4, 3, 6). Solution: The potential at P is 𝑉𝑃 = 2 −4 2 3 − 5 6 = 66 V. The electric field in space is, 𝑬 = −𝜵𝑉 = − 4𝑥𝑦 𝒂𝑥 + 2𝑥 2 𝒂𝑦 − 5𝒂𝑧 = −4𝑥𝑦 𝒂𝑥 − 2𝑥 2 𝒂𝑦 + 5𝒂𝑧 The field at P is 𝑬𝑃 = 48 𝒂𝑥 − 32 𝒂𝑦 + 5𝒂𝑧 V/m, The magnitude of 𝑬𝑃 is 𝑬𝑃 = 482 + (−32)2 +52 = 57.9 V/m Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. The direction of 𝑬𝑃 is given by the unit vector, 48 𝒂𝑥 − 32 𝒂𝑦 + 5𝒂𝑧 𝒂𝐸,𝑃 = = 0.829 𝒂𝑥 − 0.553 𝒂𝑦 + 0.086 𝒂𝑧 57.9 The displacement density is, 10−9 𝑫 = 𝜖0 𝑬 = −4𝑥𝑦 𝒂𝑥 − 2𝑥 2 𝒂𝑦 + 5𝒂𝑧 36𝜋 𝑫 = (−35.4𝑥𝑦 𝒂𝑥 − 17.71𝑥 2 𝒂𝑦 + 44.3𝒂𝑧 ) × 10−12 C/m3 The charge density is, 𝜌𝑣 = 𝜵 ∙ 𝑫 = −35.4 × 10−12 𝑦 C/m3 1 10−9 The energy density is, 𝑤𝐸 = 𝜖0 𝐸 2 = (16𝑥 2 𝑦 2 + 4𝑥 4 + 25) J/m3 2 72𝜋 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Potential and field due to dipole Dipole is a pair of charges of equal magnitude, but with opposite polarity, separated by a small distance. In the figure shown, two charges, +Q and –Q, are separated by a small distance, d, compared to the distance, r, of the field point. The potential, V, at point P is given by scalar sum of potentials at P due to charges, +Q and –Q, Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Potential and field due to dipole When r >> d, R2 ≈ R1 + d cos θ, and R1R2 = r2, hence Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Potential and field due to dipole Let us define a vector length d which starts at - Q and ends at + Q, and then define a dipole moment p = Qd ; the potential is then In general, potential due to dipole placed any where is given by, where r’ is the position vector of dipole centre, and r is the position vector of the field point. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Potential and field due to dipole Example: Given a dipole moment 𝒑 = 3𝒂𝑥 − 2𝒂𝑦 + 𝒂𝑧 nC m in free space, find potential at (2, 3, 4). Solution: The position vector of field point is 𝒓 = 2𝒂𝑥 + 3𝒂𝑦 + 4𝒂𝑧 and its magnitude is r = 4 + 9 + 16 = 5.385 , and r2 = 29. Unit vector of r is 1 𝒂𝑟 = (2𝒂𝑥 + 3𝒂𝑦 + 4𝒂𝑧 ) 5.385 The potential at (2, 3, 4) is 10−9 where 𝒑 ∙ 𝒂𝑟 = (3𝒂𝑥 − 2𝒂𝑦 + 𝒂𝑧 ) ∙ (2𝒂𝑥 + 3𝒂𝑦 + 4𝒂𝑧 ) = 0.7428 × 10−9 5.385 2 10−9 and 4𝜋𝜖0 𝑟 = 4𝜋 29 = 3.222 × 10−9 , hence V = 0.2305 volts. 36𝜋 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Continuity Equation If a current density, J, is passing through an infinitesimal surface area, dS, then the infinitesimal current is, J · dS, and the total current passing out of a closed surface is This current is result of rate of decrease of charge, Qi, inside the closed surface. By divergence theorem, we know that Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Continuity Equation If the volume inside the closed surface has a volume charge density ρv then the charge Qi can be replaced by the volume integral of charge density. Interchanging the order of differentiation and integration yields, Thus, divergence of current density is equal to the rate of decrease of charge density. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions Metal (conductor) to free space boundary: Consider a closed path, abcda; we know that line integral of electric field around a closed path is zero. Since electric field inside the conductor is zero, we need to consider the fields (both tangential and normal) in the free space only. Hence, Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions We let ∆h to approach zero, while keeping ∆w small but finite; this results in Now consider the small volume in the figure. If Q is the charge inside this volume, then by Gauss law, Integrating over the three surfaces yields, Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions Since fields inside the conductor are zero, the integration over the bottom surface is zero. As we let ∆h to approach zero, the integration over the side surface is also zero. If there exists a surface charge with density ρS then, Therefore, the boundary conditions for conductor-to-free space boundary are, the tangential components of electric field and displacement density are zero; and the normal component of displacement density is equal to the surface charge density. and Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions Example: Given the potential, 𝑉 = 100 𝑥 2 − 𝑦 2 , and a point P (2, - 1, 3), which is on conductor-free space boundary, determine potential, electric field, displacement density, and surface charge density at P. Solution: The potential at P is, 𝑉𝑃 = 100 22 − (−1)2 = 300 𝑉, and since conductor surface is an equipotential surface, the potential every where on conductor-free space boundary is 300 V. Hence equation for this boundary is 300 = 100 𝑥 2 − 𝑦 2 ֜ 𝑥2 − 𝑦2 = 3 The electric field is, 𝑬 = −𝜵𝑉, hence Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions The displacement density, D = Є0 E, Since there can not be tangential component of D at point P, the normal component of D is given by, The surface charge density at P is obtained as, Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions Dielectric to dielectric boundary: Using around the closed path shown, note that as ∆h goes to zero we obtain, Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions The tangential displacement densities across the boundary are related as, Consider the small volume shown in the figure. Applying Gauss law, the flux coming out of the closed surface is equal to the charge inside that surface. Hence, Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions However, at the boundary of two perfect dielectrics surface charges do not exist, therefore ρS is zero. Therefore, Conclusion: At the boundary of two perfect dielectrics, the tangential component of electric field is continuous, while the tangential component of displacement density is discontinuous; the normal component of displacement density is continuous, while the normal component of electric field is discontinuous. Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions Dielectric to dielectric boundary: Expressions for D2 and E2 in terms of D1 and E1 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA. Boundary conditions Dielectric to dielectric boundary: Expressions for D2 and E2 in terms of D1 and E1 Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.  Thank you Lecture notes of Dr. Raghavendra G. Kulkarni, Distinguished Professor, PES University, Bengaluru, INDIA.