Lec23 Physics 101 PDF

Lecture 23 Electrostatics II Introduction to Charges: In the last lecture, we discussed the fundamental concept of charges. Charges are entities that exert forces on each other. These forces are governed by certain principles. One key fact is that charge is c...

Lecture 23 Electrostatics II Introduction to Charges: In the last lecture, we discussed the fundamental concept of charges. Charges are entities that exert forces on each other. These forces are governed by certain principles. One key fact is that charge is conserved, meaning the total amount of charge remains constant in any closed system. Coulomb's Law: Coulomb's Law describes the force between two-point charges. If there are two charges, q1 and q2 , separated by a distance r , the force F between them is given by: q1q2 F =k r2 where k is Coulomb's constant. This force is a vector quantity, meaning it has both magnitude and direction. The direction of the force is along the line connecting the two charges. Importantly, if charge q1 exerts a force on charge q2 , charge q2 exerts an equal and opposite force on charge q1. Electric Fields: Previously we have discussed the concept of electric fields. When a charge is present in an electric field, it experiences a force. The strength of this force depends on the strength of the field. If a charge Q is placed in an electric field E , the force F acting on the charge is: F = QE Continuous Charge Distributions: In previous studies, the focus has been on point charges, which are idealized models where the entire charge is assumed to be concentrated at a single point. However, in many real-world scenarios, charges are distributed over a larger region of space rather than being concentrated at a single point. Understanding how to deal with these continuous charge distributions is crucial for accurately calculating the electric fields they produce. Summing Up Contributions from Small Elements: When dealing with continuous charge distributions, the electric field at any point in space is the result of contributions from all infinitesimal charge elements within the distribution. This is expressed as: E = E1 + E2 + E3 +  + En Here, En represents the infinitesimal electric field contribution from the n-th small charge element. Summing these contributions gives the total electric field E. Transition to Integration: Since the charge distribution is continuous, the sum of the contributions from all infinitesimal charge elements is represented by an integral: E =  En This summation can be approximated by an integral for continuous distributions: E =  dE Vector Components of the Electric Field: The electric field E is a vector quantity, meaning it has both magnitude and direction. It can be decomposed into its Cartesian components along the x, y, and z axes: E = Exiˆ + E y ˆj + Ez kˆ Where: Ex , E y and Ez are the components of the electric field in the x, y, and z directions, respectively.  iˆ, ˆj , and kˆ are the unit vectors in the x, y, and z directions. Calculating Each Component: To find the total electric field, you need to integrate the contributions to each component over the entire charge distribution: Ex =  dEx E y =  dE y Ez =  dEz These integrals take into account the continuous distribution of charge and calculate the contributions to the electric field in each direction. Linear Charge Distribution: Consider a line of charge where charges are continuously distributed along its length. To find the electric field at a point P due to this line of charge, we break the line into small segments and sum the contributions of each segment. Suppose we have a line of charge with charge density  , which is charge per unit length. If we take a small segment of length ds , the charge dq on this segment is: dq = ds By summing the contributions of all these small segments, we can determine the total electric field at point P. Surface and Volume Charge Distributions: In two dimensions, we consider surface charge distributions. Here, charges are spread over a surface, such as a sheet of paper. The charge density  is charge per unit area. For a small area element dA , the charge dq is: dq =  dA In three dimensions, we consider volume charge distributions. Here, charges are distributed throughout a volume. The charge density  is charge per unit volume. For a small volume element dV , the charge dq is: dq =  dV Calculating Electric Fields for Different Distributions: To find the electric field due to a continuous charge distribution, we integrate over the entire distribution. For a line charge, the electric field E at a point can be calculated using: kdq E= rˆ r2 where dq is the charge element, r is the distance from the charge element to the point, and r̂ is the unit vector pointing from the charge element to the point. Electric Field Due to a Ring of Charge: Now, consider a ring of charge with radius R. We want to find the electric field along the axis of the ring. By symmetry, we only need to consider the z-component of the electric field. If we break the ring into small segments, each segment contributes a small electric field dE : Fig 23. 1: The electric field of a charged ring points along the ring axis, since field components perpendicular to the axis cancel in pairs. Starting with an infinitesimal electric field element dE : 1  ds  ds dE = = 4 0 r 2 ( 4 0 z 2 + R 2 ) This expression gives the infinitesimal electric field dE due to a small charge segment ds of the ring, where:  is the linear charge density (charge per unit length). ds is the infinitesimal segment length of the ring. r is the distance from the charge segment to the point where the field is being calculated.  0 is the permittivity of free space. The distance r is given by r = z 2 + R 2 , where z is the distance along the axis from the centre of the ring to the point of interest, and R is the radius of the ring. Component of dE along the z-axis dE z : dEz = dE cos  Here, cos  is the cosine of the angle between the electric field vector and the z-axis. This angle  can be calculated using the geometry of the ring. Expression for cos  : z z cos  = = ( ) 1/2 r z + R2 2 This step uses the definition of cos  in terms of the z-component and the hypotenuse r. Substituting cos  into dE z : z ds dEz = ( ) 3/2 4 0 z 2 + R 2 This equation represents the infinitesimal component of the electric field along the z-axis due to a small segment ds of the charged ring. Integrating dE z around the ring: z Ez =  ds ( ) 3/2 4 0 z 2 + R 2 Since the ring is symmetrical, we integrate dE z over the entire circumference of the ring, where  ds = 2 R , the circumference of the ring. Solving the integral: z ( 2 R ) Ez = ( ) 3/2 4 0 z 2 + R 2 Simplifying, we obtain: qz Ez = ( ) 3/2 4 0 z 2 + R 2 Here, q =  (2 R) is the total charge of the ring. Electric field for z  R : When the point of interest is far along the z-axis z  R , the expression simplifies. Under this approximation: qz q Ez  = 4 0 z 3 4 0 z 2 This is the electric field along the axis of the ring when the observation point is far from the ring. It resembles the electric field of a point charge q at a distance z. the steps provide the derivation of the electric field along the axis of a charged ring, starting from the contribution of a small segment and integrating over the entire ring, leading to a simplified form for distances much larger than the ring radius. Electric Field Due to a Disk of Charge: Consider a uniformly charged disk with charge density . To find the electric field at a point along the axis of the disk, we divide the disk into concentric rings and integrate over these rings: Fig 23. 2: A uniformly charged disk of radius R. The field on the z-axis due to an infinitesimally small ring element at a radius  with thickness d. To find the electric field at a point along the axis of a uniformly charged disk, we divide the disk into concentric rings and integrate over these rings. Let's break down the given steps in detail: Surface Charge Density and Elemental Charge: The disk has a uniform surface charge density  , which is the charge per unit area. Consider an infinitesimally small ring element at a radius  with thickness d. The area of this ring element dA is given by the circumference of the ring ( 2 ) times the thickness ( d ). dA = 2d The charge dq on this ring element is:  dq =  dA =  ( 2 ) d Electric Field Contribution by the Ring Element: To find the electric field dE z at a point along the axis (distance z from the disk centre) due to this ring element, we use Coulomb's law. The vertical component dE z of the electric field due to the ring element is: z 2 d  dEz = ( ) 3/2 4 0 z 2 +  2 Simplifying the expression: z 2 ( z +  2 ) ( 2 ) d −3/2 dEz = 4 0 Integrating Over All Rings: To find the total electric field Ez along the axis, integrate dE z from  = 0 to  = R (the radius of the disk): z R (z ) ( 2 ) d −3/2 Ez =  dEz = 2 + 2 4 0 0 Solving the Integral: Evaluate the integral: R 2 d  0 ( z +  2 )3/ 2 2 Using a suitable substitution (  = z tan  ), this integral evaluates to: 2 d ( ) R 2 2 = 2  z2 + 2  = 2 R  0 (z +  ) 2 2 3/2 z   0 z z 2 + R2 − z Final Electric Field Expression: Plugging the evaluated integral back: Ez = z 2  4 0 z 2 ( z 2 + R2 − z =)   1 − 2 2 0  z   z + R2  Special Cases: 1. When z  R : If z  R , the distance z is much smaller than the radius R , implying the point is very close to the disk:   z  Ez = 1 − 2  2 0  z + R2   Ez = 2 0 This is the same as the electric field near an infinite sheet of charge. 2. When z  R : If z  R , the distance z is much larger than the radius R , implying the point is far from the disk:  R2 Ez = 4 0 z 2 Since q =  R 2 (total charge on the disk): q Ez = 4 0 z 2 This is the same as the electric field due to a point charge q at a distance z.This process involves breaking the disk into infinitesimal rings, calculating the electric field contribution from each ring, and integrating these contributions to find the total electric field along the axis of the disk. The results match known cases for very close and very far distances, providing confidence in the calculations. Electric Field Due to infinite Line of Charge: To calculate the electric field due to an infinite line of charge, consider a line of charge distributed along the z-axis. We want to find the electric field at a point P on the y-axis. Here's a step-by-step explanation of the process: Fig 23. 3: The field of a charged line is the vector sum of the fields from all the individual charge elements dq along the line. Linear Charge Density: Let the linear charge density be  , which is the charge per unit length. Electric Field Contribution by an Elemental Segment: Consider a small segment of the line of charge dz at a distance z from the origin. The charge dq on this segment is: dq = dz The distance r from this segment to the point on the y-axis at distance y is given by: r= y2 + z2 The electric field dE due to this charge segment at the point is: 1 dq 1  dz dE = = 4 0 r 2 4 0 y + z 2 2 Components of the Electric Field: The electric field dE can be decomposed into two components: dE y along the y-axis and dE z along the z-axis. Due to symmetry, the z -components from opposite segments cancel out, leaving only the y -components. The y -component dE y is: 1  dz y 1  ydz dEy = dE cos  = = 4 0 y + z 2 2 y2 + z2 4 0 ( y + z 2 )3/2 2 Integration Over the Entire Line: To find the total electric field E y , integrate dE y from z = − to z =  : z = y z = dz Ey =  dE y =  z =− 4 0 z =− ( y + z 2 )3/2 2 Simplifying the Integral: Use the substitution z = y tan  , hence dz = y sec2  d : y  /2 y sec 2  d y  /2 Ey = 4 0  − /2 y sec  3 3 = 4 0 y 2  − /2 cos d Simplifying further:    sin  − /2 = (1 − (−1) ) =  /2 Ey = 4 0 y 4 0 y 2 0 y Final Result and Symmetry: The electric field due to an infinite line of charge at a point on the y-axis is given by:   E= 2 0 y Given the cylindrical symmetry of the problem, this result can be generalized to any point at a perpendicular distance r from the line of charge:  E= 2 0 r By breaking the infinite line of charge into infinitesimal segments, calculating their contributions to the electric field, and integrating over the entire length of the line, we find that the electric field at a perpendicular distance r from an infinite line of charge is inversely proportional to the distance r and depends on the linear charge density . Corona Discharge: Corona discharge is an electrical discharge brought about by the ionization of a fluid such as air surrounding a conductor that is electrically charged. It occurs when the electric field around the conductor becomes sufficiently strong to ionize the surrounding air or other gaseous medium, but not strong enough to cause a complete electrical breakdown or spark. Conditions for Corona Discharge: High Voltage: The conductor must be at a high voltage relative to its surroundings. Sharp Points or Edges: Corona discharge is more likely to occur at sharp points or edges on the conductor because the electric field is concentrated in these areas.  Gas Type and Pressure: The surrounding gas and its pressure influence the onset of corona discharge. Air is the most common medium, but other gases like SF₆ (sulphur hexafluoride) are used in high-voltage equipment to suppress corona. Fig 23. 4: The hissing noise at high voltage transmission lines is due to corona discharge. Corona discharge occurs due to high voltage which causes surrounding medium (air) to ionized. Electric Flux: Electric flux is a measure of the number of electric field lines passing through a surface. If we have an electric field E passing through a surface area A at an angle  , the flux  E is given by: Fig 23. 5: Field lines for a uniform electric field through an area A for different angles to the field. Because the number of lines that go through the shaded area A is the same as the number that go through A, we conclude that the flux through A is equal to the flux through A and is given by  E = EA cos .  E = EA cos  For a curved surface, we sum the flux through small area elements dA :   E =  E  dA Fig 23. 6: Flux for a curved surface will be the sum of the flux through small area elements. Gauss's Law: Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is one of Maxwell's four equations, which form the foundation of classical electromagnetism. Gauss's Law states that the net electric flux through any closed surface is equal to the charge enclosed divided by the permittivity of the medium. Mathematically, it is expressed as: Qenclosed  E =  E. dA = 0 where:  E. dA is the surface integral of the electric field E over a closed surface S. dA is a vector representing an infinitesimal area on the surface S , pointing outward. Qenclosed is the total electric charge enclosed within the surface S.  0 is the permittivity of free space (vacuum permittivity). Explanation: Gauss's Law essentially states that the total electric flux through a closed surface is proportional to the total charge enclosed within that surface. The electric flux is a measure of the number of electric field lines passing through a given area. If more charge is enclosed, more field lines emanate from or terminate at the charges, resulting in greater flux. 1. Electric Field of a Point Charge: Consider a point charge q located at the origin. The electric field E at a distance r from the charge is given by Coulomb's Law: q E= r 4 0 r 2 where r is the unit vector pointing radially outward from the charge. 2. Flux Through a Spherical Surface: Imagine a spherical surface of radius r cantered around the charge. The surface area of the sphere is 4 r 2. The electric field is radially symmetric and has the same magnitude at all points on the surface. The flux  E through the spherical surface is:  E =  E. dA Since E and dA are parallel and | E | is constant over the sphere, this simplifies to:  E = E  dA = E  4 r 2 q Substituting E = : 4 0 r 2  q  q E =  2   4 r 2 =  4 0 r  0 3. Generalization: For a general distribution of charges, consider a closed surface S enclosing a total charge Qenclosed. By the superposition principle, the total flux through S is the sum of the fluxes due to each individual charge. Hence:  Qenclosed  E =  E  dA = 0 This concludes the derivation, showing that the electric flux through any closed surface is proportional to the enclosed charge, as stated by Gauss's Law. Applications: 1. Symmetry: Gauss's Law is particularly powerful for calculating electric fields of symmetric charge distributions (e.g., spherical, cylindrical, or planar symmetry). 2. Conductors: Helps determine the distribution of charge on conductors and the resulting electric field. 3. Capacitors: Used to analyse the electric field and potential in capacitors. 4. Electrostatics: Fundamental in solving electrostatic problems and understanding the behaviour of electric fields in various configurations. Electric Field Due to infinite Parallel Plates: application of Gauss's Law in finding the electric field between two infinite parallel plates with uniform surface charge densities: Fig 23. 7: Electric field E between two infinite parallel plates with uniform surface charge densities. A cylindrical Gaussian surface surrounds a parallel plate. 1. Electric Flux Definition:  =  E  dA  This integral defines the electric flux  through a closed surface, where E is the electric field and dA is a differential area vector on the surface. 2. Decomposing the Flux: =  outer E  dA +  inner E  dA +  side E  dA cap cap walls This equation breaks down the total electric flux through the Gaussian surface into three parts: The flux through the outer cap. The flux through the inner cap. The flux through the side walls. 3. Evaluating Each Component:  = 0 + EA + 0 For the chosen Gaussian surface (a cylindrical "surface" that straddles one of the plates): The flux through the outer cap is zero because the electric field outside the plates cancels out. The flux through the inner cap is EA because the electric field E is perpendicular to the surface and uniform over the area A of the cap. The flux through the side walls is zero because the electric field is parallel to these surfaces and doesn't contribute to the flux. 4. Relating Flux to Charge:  0 = q According to Gauss's Law, the electric flux  through a closed surface is equal to the enclosed charge q divided by the permittivity of free space  0. 5. Enclosed Charge in Terms of Surface Charge Density:   0 EA =  A The charge q enclosed by the Gaussian surface is the surface charge density  times the area A of the inner cap. 6. Solving for the Electric Field:  E= 0 Simplifying this equation gives the electric field E between the plates. Electric Field Due to a long, straight wire: Consider a straight wire with a constant charge distributed along its length with charge density  , we use a cylindrical Gaussian surface with radius r and length L. Fig 23. 8: A cylindrical Gaussian surface surrounds a line charge. 1. Gaussian Surface: We create an imaginary cylindrical Gaussian surface around the wire to simplify our calculations. The surface is closed and we can choose its shape to our advantage. 2. Calculating the Electric Flux: o The flux through the cylindrical surface is the product of the electric field and the surface area of the cylinder. 1. Gauss's Law:   0  E  dA = q This is the integral form of Gauss's Law, where  0 is the permittivity of free space, E is the electric field, dA is a differential area element, and q is the charge enclosed by the Gaussian surface. 2. Choosing the Gaussian Surface: For a long, straight wire with a uniform linear charge density  (charge per unit length), we choose a cylindrical Gaussian surface with: o Radius r o Height h This surface is coaxial with the wire. 3. Calculating the Electric Flux: The flux through the Gaussian surface is:  0  E  dA Since the electric field E is radial and perpendicular to the surface of the cylinder, and is uniform at a distance r from the wire, this simplifies to:  0 E ( 2 rh ) Here, 2 r is the circumference of the cylinder's base, and h is the height of the cylinder. 4. Enclosed Charge: The charge enclosed by the Gaussian surface is: q = h Here,  is the linear charge density, and h is the height of the cylinder. 5. Applying Gauss's Law: According to Gauss's Law:  0 E ( 2 rh ) = h Simplifying this equation to solve for E :  E= 2 0 r For a point charge q at the centre of a sphere of radius r , the electric field is radial and has the same magnitude at every point on the surface. The flux through the sphere is: q  E  dA = E  4 r = 2 0 Solving for E , we get: 1 q E= 4 0 r2 This result is consistent with Coulomb's Law. Example.3: Electric Field Due to Infinite Sheet of Charge: Fig 23. 9: A cylindrical Gaussian surface penetrating an infinite sheet of charge. (b) A cross section of the same Gaussian cylinder. The flux through each end of the Gaussian surface is EA. There is no flux through the cylindrical surface. We use Gauss's Law to determine the electric field near an infinite plane of charge. 1. Gauss's Law:  0  E  dA = q This is the integral form of Gauss's Law, where  0 is the permittivity of free space, E is the electric field, dA is a differential area element on the Gaussian surface, and q is the charge enclosed by the surface.  2. Choosing the Gaussian Surface: For an infinite plane of charge with a surface charge  (charge per unit area), we choose a Gaussian surface in the shape of a rectangular box (a "surface") that extends symmetrically on both sides of the plane. 3. Calculating the Electric Flux: The total electric flux through the Gaussian surface is the sum of the flux through the two caps (top and bottom) of the surface.  0 ( EA + EA) Since the electric field E is perpendicular to the plane and uniform over each cap with area A , the flux through each cap is EA. There are two such caps, so the total flux is EA + EA. 4. Enclosed Charge: The charge enclosed by the Gaussian surface is: q = A Here,  is the surface charge density, and A is the area of the plane enclosed by the Gaussian surface. 5. Applying Gauss's Law: According to Gauss's Law:  0 ( EA + EA) =  A Simplifying this equation to solve for E : 2 0 EA =  A  E= 2 0 This derivation shows how Gauss's Law can be used to find the electric field near an infinite sheet of charge. The result is that the electric field is uniform and directed perpendicular to the  plane, with a magnitude given by. 2 0 Electric Field Due to a Spherical Shell: We consider a spherical shell with charge distributed only on its surface and calculate the electric field both inside and outside the shell using Gauss's Law. Fig 23. 10: A spherical Gaussian surface surrounds a spherical shell of radius R and charge q. 1. Inside the Shell: Gauss's Law: Gauss's Law states that the total electric flux through a closed surface is proportional to the charge enclosed by that surface. qenc  E  dA = 0 Applying Gauss's Law Inside the Shell: For a point inside the spherical shell (at radius r where r  R , with R being the radius of the shell), consider a Gaussian surface that is a sphere of radius r. Since the charge is distributed only on the surface, the enclosed charge qenc within this Gaussian surface is zero. Fig 23. 11: A spherical Gaussian surface that is a radius r inside a spherical shell of radius R and charge q. qenc = 0 Therefore, the electric flux is zero:  E  dA = E  4 r =0 2 This implies that the electric field inside the shell is zero: E =0 ( spherical shell, r  R ) 2. Outside the Shell Applying Gauss's Law Outside the Shell: For a point outside the spherical shell (at radius r where r  R ), consider a Gaussian surface that is a sphere of radius r. The enclosed charge qenc is the total charge q on the spherical shell. Fig 23. 12: A spherical Gaussian surface that is a radius r surrounds a spherical shell of radius R and charge q. qenc = q According to Gauss's Law, the electric flux through this surface is: q  E  dA = 4 r E =  2 0 Solving for the electric field E : 1 q E= 4 0 r 2 Thus, the electric field outside the shell behaves as if all the charge were concentrated at the centre of the shell: 1 q E= ( spherical shell, r  R ) 4 0 r 2 This result shows the electric field due to a spherical shell of charge follows the same pattern as the electric field due to a point charge for points outside the shell, and is zero inside the shell. Conductors and Electric Fields: In conductors, the electric field inside is zero because the charges redistribute to cancel any internal fields. When a charge is placed inside a conducting cavity, the charges on the conductor rearrange to neutralize the effect, resulting in no electric field within the cavity. 1. Conducting Sphere: o If a charge is placed inside a hollow conducting sphere, the charges on the inner surface of the conductor will rearrange to cancel the field within the cavity. 2. Conductor Surface: o If a charge is placed on a conductor, it will move to the surface to minimize the repulsive forces between like charges. The electric field just outside the surface of a conductor is perpendicular to the surface and proportional to the surface charge density. Discontinuity of the Electric Field at a Charged Surface: When there is a surface charge density  on a boundary, the electric field experiences a discontinuity as you move across the surface. Gauss's Law can be used to quantify this discontinuity. Applying Gauss's Law to a Surface Charge: Consider a small Gaussian surface that straddles the surface with area A. The surface has one face just above the surface and one face just below the surface. The electric field on each side of the surface is represented as En1 (above the surface) and En 2 (below the surface). Fig 23. 13: An infinite sheet of charge, with a Gaussian surface straddling the sheet. Gauss's Law states: qenc  E  dA = 0 For the surface, the only contribution to the flux is through the flat faces of the surface (since the sides are perpendicular to the electric field and contribute nothing to the flux). The charge enclosed qenc is the surface charge density  times the area A. qenc =  A The electric flux through the surface is the sum of the flux through the top and bottom faces:  E  dA = E n2 A − En1 A By Gauss's Law: A En 2 A − En1 A = 0 Dividing by the area A :  En 2 − En1 = 0 This equation tells us that the difference between the electric field just outside ( En1 ) and just inside ( En 2 ) a charged surface is directly proportional to the surface charge density  , with 1 the proportionality constant being. 0 Potential Energy and Electric Fields: Fig 23. 14: Electric field (E) between two infinite parallel plates with uniform surface charge densities (+σ and −σ). A charge 𝑞 moves from the positively charged plate to the negatively charged plate, experiencing a force 𝐹 due to the electric field. Electric potential energy is the energy stored in an electric field due to the positions of charges. Moving a charge against an electric field requires work, which increases the potential energy of the system. This concept is analogous to gravitational potential energy. 1. Relationship Between Potential Energy and Work: The change in electric potential energy ( U ) between two points A and B is related to the work done ( Wab ) to move a charge from point A to point B: U B − U A = Wab U = −Wab 2. Work Done by Electrostatic Force: The work done by an electrostatic force F when moving a charge from point A to point B is given by: b U b − U a = −  F  ds a 3. Electrostatic Force and Potential Energy: The electrostatic force is conservative and can be represented by a potential energy function. Comparing with gravitational force: m1m2 1 q1q2 F =G and F= r2 4 0 r 2 Where G is the gravitational constant and  0 is the permittivity of free space. 4. Electric Field and Force: The relationship between the electric field E and the force F on a charge q is: F E= q For a gravitational field g and a mass m : F g= m 5. Calculation of Potential Energy: Consider a charge q2 in the electric field created by another charge q1. The work done to move q2 from distance ra to rb is: Fig 23. 15: Electric field E experiencing by charge q2 due to charge q1 at a distance r. rb U b − U a = −q2  Ex dr ra For a point charge, the electric field E due to q1 at a distance r is: 1 q1 E= 4 0 r 2 Thus, rb 1 dr Ub − U a = − q1q2  4 0 ra r2 1 1 1 Ub − U a = q1q2  −  4 0  rb ra  6. Reference Point and Potential Energy: If we take the reference point ra →  where U a → 0 , then at a distance rb = r , the potential energy Ub = U ( r ) is: 1 q1q2 U (r ) = 4 0 r 7. Comparison with Gravitational Potential Energy:  The electric potential energy U ( r ) is analogous to the gravitational potential energy: 1 q1q2 U electric (r ) = 4 0 r m1m2 U gravitational (r ) = −G r Electric potential energy is the energy stored due to the relative positions of charges. Moving a charge in an electric field requires work, changing the potential energy. This work is path- independent because the electrostatic force is conservative. The potential energy between two- point charges is inversely proportional to the distance between them, analogous to gravitational potential energy. Summary In this lecture, we have explored the foundational concepts of electrostatics, including charges, electric fields, continuous charge distributions, and Gauss's Law. We have seen how these principles are applied in real-world problems and practical applications. Mastery of these topics is essential for a deeper understanding of electromagnetism and its applications in technology and nature.