Electric Potential PDF

# Chapter V - Electric Potential ## Introduction - Previous chapters discussed various methods for calculating electric fields. - Chapters 3 and 4 addressed Coulomb's law and Gauss's law. - This chapter introduces a new quantity called electric potential, which is a scalar quantity unlike electric fields and forces. - The relationship between electric potential and electric field is established. - Electric potential allows for an easier method of finding electric fields by means of differentiation, rather than integration. - The concept of electric potential is more useful in practical circuit analysis and is commonly known as voltage. - The chapter also discusses the conservative nature of electrostatic fields, which leads to the derivation of another of Maxwell's equations. ## Work Done by an Electric Field - A charged particle placed in an electric field experiences a force F=QE. - This force causes the charge to move in a certain direction, defined by the field and the polarity of the charge. - Work is done on the charge when it is displaced due to the action of the force. - Work done on the charge is calculated by multiplying the component of the force parallel to the displacement with the displacement, resulting in: W = F cos(theta) * d where: - **W** is work done - **F** is force - **d** is displacement - **theta** is the angle between force and displacement - Work done can be expressed as a dot product: W = F · d - If force is not constant, the work can be calculated by performing integration along the path: W = ∫F · dl - Where **dl** is an infinitesimal displacement vector tangent to the path - Integration is performed from the initial point to the final point. - If the electric field E is uniform, a constant force F is exerted on a charge Q, causing a displacement of d. - The work done by the electric field is calculated by: W = QEd - The charge and the field define a charge-field system. - Work done causes a transfer of energy. - Change in electric potential energy is equal to the negative of the work done: ΔU = -W - For a uniform electric field, the change in potential energy is: ΔU = -QEd - The work done by the electric field is internal to the system. - To move the charge against the electric field, external energy input is required, which is equal to the change in potential energy: W = ΔU. - The work done to move a charge in an electric field is defined by the line integral: W = -Q∫E · dl - Where integration is performed between two points in the electric field. - A positive work value indicates an increase in the charge-field system energy. A negative work value indicates a decrease in the system energy. ## Illustrative Problem 5.1 A 100-µC charge must be transferred along the arc of a circle x² + y² = 25 from point A(5, 0, 0) to B(3, 4, 0) in an electric field given by E = yax + xay + 5az. Calculate the work done. **Solution:** The electric field is nonuniform, so integration is required. W = -100(10-6) ∫AB(yax + xay + 5az) · (dxax + dyay + dzaz) Simplifying the dot product: W = -100(10-6) ∫AB(ydx + xdy + 5dz). Using the linearity of the integral: W = -100(10-6)[∫ABydx + ∫ABxdy + ∫AB5dz]. The third integral vanishes because the lower and upper limits are the same. Using the equation of the circle, y can be expressed in terms of x and x can be expressed in terms of y for the first and second integrals respectively. W = -100(10-6)[∫AB√(25 - x²)dx + ∫AB√(25 - y²)dy]. Evaluation of the integrals yields a work done of -1.20 mJ. ## Illustrative Problem 5.2 Calculate the work done in transferring the same charge from A(5, 0, 0) to B(3, 4, 0) along a straight-line path, given the same electric field from the previous problem. **Solution:** First, we find the equation of the line passing through A(5, 0, 0) and B(3, 4, 0). The slope is -2, and the equation is given by: y = -2x + 10. The work done can be calculated as follows: W = -100(10-6)∫53(-2x + 10)dx = -1.20 mJ The work done is the same in both cases. This is because electrostatic fields are conservative, meaning the work done is independent of the path taken. ## The Electric Potential Difference - The electric potential difference (VAB) is defined as the change in electric potential energy per unit charge when moving a positive test charge from point B to point A in an electric field. - It is calculated as: VAB = -∫BAE · dl - This definition is also equivalent to the negative of the work done per unit charge. - Potential difference can be expressed in terms of work as well: W = Q · ΔV - Where **ΔV** is the potential difference. - The unit of electric potential difference is the Volt (V), defined as 1 J/C. - Potential difference is often referred to as voltage. - To measure the absolute potential (VA) at point A, a reference point B is chosen with zero potential (VB = 0), resulting in: VA = VAB. - Electric potential at a particular point is defined as the work done per unit charge to bring a positive test charge from a point of zero potential to that point. - In practical circuits, the ground is typically used as the reference point. - In mathematical analysis, infinity is typically used as the reference point. - This results in the following expression for electric potential at a point: V = -∫∞AE · dl - The potential due to a point charge is inversely proportional to the distance from the charge. - An equipotential surface is defined as a surface where the potential is the same at all points. - Electric field lines are always orthogonal to equipotential surfaces. - The work done in moving a charge along an equipotential surface is always zero because the displacement and the electric field are perpendicular. ## Illustrative Problem 5.3 Derive an expression for finding the potential difference between two points A and B due to a point charge Q. **Solution:** The electric field due to a point charge is E = (Q/(4πεor²))ar. Using spherical coordinates, the potential difference can be calculated using the following integral: VAB = -∫BAE · dl = -∫BA(Q/(4πεor²))ar · drar = -∫BA(Q/(4πεor²))dr = Q/(4πεο)[(1/rA) - (1/rB)]. ## Illustrative Problem 5.4 Find the electric potential at point A, a distance r from a point charge Q. **Solution:** To find the electric potential at A, it is convenient to use a reference point B where the potential is zero. Setting B to infinity and leveraging the result from the previous illustrative problem, the potential at A is: VA = VAB = Q/(4πεor) ## Illustrative Problem 5.5 A point charge of 120 nC is located at (2, -3, 5). Find the electric potential at (-4, 0, 2). **Solution:** The distance between the point charge and the point (-4, 0, 2) is calculated as r =√[(-4 - 2)² + (0 - (-3))² + (2 - 5)²] = √54. The electric potential is then calculated as: V = kQ/r = (9 * 10⁹)(120 * 10⁻⁹)/√54 = 146.97 V. ## Illustrative Problem 5.6 A point charge of 10 nC is located at the origin. If the potential at point (0, 5, 4) is 1 V, find: (a) the potential at A(-2, 0, -1), (b) the potential at B(3, -2, 0), and (c) the potential difference VAB. **Solution:** We will use point P(0, 5, 4) as the reference point with VP = 2 V. **(a)** The potential at A is calculated as follows: VA = VP + VAP = VP + (9)(-10⁻⁹)[(1/√5) - (1/√41)] = 26.19 + 2 = 28.19 V. **(b)** Similarly, the potential at B is calculated as follows: VB = VP + VBP = VP + (9)(-10⁻⁹)[(1/√13) - (1/√41)] = 10.91 + 2 = 12.91 V. **(c)** The potential difference VAB is calculated as follows: VAB = VA - VB = 28.19 - 12.91 = 15.28 V ## Illustrative Problem 5.7 Determine the potential difference VAB if a point charge Q is displaced radially from p = B to p = A from an infinite line charge having a charge density pL. **Solution:** The potential difference can be found from the definition of potential difference: VAB = -∫BAE · dl = -∫BA(pL/(2πεοr))ar · drar = -∫BA(pL/(2πεοr))dr = (pL/(2πεο))ln(pA/pB) ## Illustrative Problem 5.8 The electric field in a region of free space is given by E = (4x² - 1)ax + 3yay - 5az. Find: (a) Vxy if X(5, -2, 0) and Y(1, 4, -2), (b) Vx if V = 0 at A(6, -2, 4), and (c) Vy if V = -3 V at B(-1, 0, 2) **Solution:** **(a)** Vxy can be calculated using the line integral: Vxy = -∫YXE · dl = -∫YX[(4x² - 1)ax + 3yay - 5az] · (dxax + dyay + dzaz) = -∫YX[(4x² - 1)dx + 3ydy - 5dz] = -133.33 V. **(b)** To find Vx, we first calculate VXA: VXA = -∫AXE · dl = -∫AX[(4x² - 1)ax + 3yay - 5az] · (dxax + dyay + dzaz) = -∫AX[(4x² - 1)dx + 3ydy - 5dz] = 100.33 V. Since V = 0 at A, then VxA = Vx = 100.33 V. **(c)** To find Vy, we first calculate VYB: VYB = -∫BYE · dl = -∫BY[(4x² - 1)ax + 3yay - 5az] · (dxax + dyay + dzaz) = -∫BY[(4x² - 1)dx + 3ydy - 5dz] = -44.67 V. Therefore, Vy = VYB + VB = -44.67 + (-3) = -47.67 V. ## Electric Potential Due to Continuous Charge Distributions and a System of Charges - The electric potential due to a point charge is V = Q/(4πεοr). - To find the electric potential due to a continuous charge distribution, we can integrate the potential due to each differential charge element dQ: V(r) = ∫(dQ/(4πεοr)) - Where **r** is the distance from the differential charge element to the point where the potential is being measured. - Equation for dQ: dQ = pLdl (line charge), dQ = psds (surface charge), dQ = pdv (volume charge) - Using the superposition principle, the electric potential due to a system of charges is simply the sum of the potentials due to each individual charge. ## Illustrative Problem 5.9 A tetrahedron is formed by four equilateral triangles, with the centroid of the base located at the origin. Point charges with equal magnitude of 24 nC are located at the corners of the base. Each side of the triangle measures 6 meters. Determine the point charge that should be placed at the origin so that the potential at the apex of the pyramid due to all five point charges is zero. **Solution:** Let Q1 = 24 nC, r1 = 6 meters, r2 = 2√6 meters, and Q2 be the unknown point charge. The potential at the apex of the pyramid due to all five point charges should be zero. Setting this up as an equation: 3[kQ1/r1] + kQ2/r2 = 0. Solving for Q2, we get: Q2 = -3(Q1/r1)(r2) = -3(24 * 10⁻⁹ / 6)(2√6) = -58.79nC ## Illustrative Problem 5.10 A -50 nC point charge is located at point Q(3, -4, 0). A uniform line charge of 80 nC/m is located at L(-6, 8). Find: (a) the potential at point A(6, 2, 1) if the potential at the origin O is zero, (b) the potential at point B(-2, 4, 3) if the potential at (1, 0, 2) is 25 V, and (c) VAB if the potential at the origin is -10 V **Solution:** **(a)** The total potential at A is the sum of the potentials due to Q and the line charge. Since the potential at the origin is zero: VAO = (9 * 10⁹)(-50 * 10⁻⁹)[(1/√46) + (1/ 5)] = 23.65 V For the line charge: VAO = (18 * 10⁹)(80 * 10⁻⁹)ln(10/6√5) = -423.21 V. Therefore, the total potential at A is: VA = VAO + VAO = 23.65 + (-423.21) = -399.56 V **(b)** Let X be the point (1, 0, 2). Vx = 25 V. Using the point charge and the line charge, we get: Vax = (9 * 10⁹)(-50 * 10⁻⁹)[(1/2√6) - (1/7√2)] = 46.40 V. VBX = VB - VX, so VB = 46.40 + 25 = 71.40 V. For the line charge: VBX = (18 * 10⁹)(80 * 10⁻⁹)ln(√113/4√2) = 908.39 V. Therefore, the total potential at B is: VB = 71.40 + 908.39 = 1004.79 V. **(c)** The potential difference VAB is equal to the sum of the potential differences due to Q and the line charge. For the point charge: VAO = 23.65 V (from part a) VA = VAO + VO = 13.65 V VBO = (9 * 10⁹)(-50 * 10⁻⁹)[(1/√2) - (1/5)] = 44.54 V VB = VBO + VO = 34.54 V For the line charge: VAD = -423.21 V (from part a) VA = VAO + VO = -433.21 V. VBO = (18 * 10⁹)(80 * 10⁻⁹)ln(10/4√2) = 820.39 V VB = VBO + VO = 810.39 V Therefore, the total potential at A is -419.56 V, and the total potential at B is 844.93 V. The potential difference VAB is: VAB = VA - VB = -419.56 - 844.93 = -1264.49 V ## The Conservative Nature of Electrostatic Fields - Electrostatic fields are conservative. - The work done in transferring a charge between two points in an electrostatic field is independent of the path taken. - The work done moving a charge along a closed path in an electrostatic field is always zero. - This can be expressed mathematically as: W = -Q∫E · dl = 0 - Dividing both sides by -Q: ∫E · dl = 0 - This equation is a consequence of Stokes' theorem, which states that the line integral of a vector field around a closed path is equal to the integral of the curl of the same vector through the surface defined by the closed path. - Applying Stokes' theorem results in the equation: ∫E · dl = ∫(∇ × E) · dS = 0. - This indicates that the curl of an electrostatic field is zero (∇ × E = 0). - This equation is one of Maxwell's equations for static electric fields. - It implies that electrostatic fields are conservative. - It is also the field equivalent of KVL (Kirchhoff's Voltage Law) in electrical circuit analysis. - Electric fields that vary with time are not conservative. - Irrotational fields have zero curl and are often referred to as conservative fields. ## The Potential Gradient - If the curl of a vector field is zero, it can be expressed as the gradient of a scalar function: ∇ × Vf = 0. - This implies that the electric field E is the gradient of the electric potential V: E = -∇V - Electric potential is a scalar function whose gradient is the electric field. - This relationship can be used to calculate the electric field from the electric potential or vice versa. - The gradient of a scalar function measures the rate of change of that function with respect to distance. - The gradient is always a vector quantity. ## Illustrative Example 5.11 The electric potential field in a region is given by V = 4x²y - 3xy + 5z. Find: (a) the potential at point P(-2, 1, 4), (b) the electric field, its magnitude and direction, (c) the flux density, and (d) the charge density at point P. **Solution:** **(a)** The potential at point P is: VP = 4(-2)²(1) - 3(-2)(1) + 5(4) = 42 V **(b)** The electric field is calculated as the negative gradient of the potential function: E = -∇V = -[(8xy - 3y)ax + (4x² - 3x)ay + 5az]. At point P, E = 19ax - 22ay + 5az. Its magnitude is √(19² + (-22)² + 5²) = 29.5 V/m. The direction of E is given by: AE = (19ax - 22ay + 5az) / 29.5 = 0.644ax - 0.7459ay + 0.1695az. **(c)** The electric flux density is given by D = εοE: D = εο(19ax - 22ay + 5az) **(d)** The charge density is equal to the divergence of D: pv = ∇ · D = ∇ · εο(19ax - 22ay + 5az) = -8εοy C/m³ At point P, pv = -8εο(1) = -8εο C/m³. ## Chapter Review **Part 1 - Multiple Choice** 1. The work done in moving a charge around a closed path in an electrostatic field is always zero. 2. The integral of the dot product of a vector with dl will give you the line integral of the vector. 3. If VAB is negative, the potential at B is higher than A. 4. The electric potential inside the conductor is constant in electrostatic equilibrium. 5. The tangential component of the electric field is always continuous at the interface of the conductor and free-space. 6. Electric potential difference is defined as the change in potential energy per unit charge. 7. Electric field is a measure of the rate of change of the electric potential with respect to position. 8. Electric potential depends on the magnitude of electric charge. **Part 2 - Problems** 1. **Charged Ring:** Derive an expression for the electric potential at any point on the axis of a charged ring with a uniform charge distribution. 2. **Infinite Line Charge:** Derive an expression for finding the potential field of an infinite line charge with a charge density at a distance r from it. 3. **Potential Function:** Given the potential V = (10/r²) sin θ cos φ, find the electric flux density at A(2, π/2, 0), and calculate the work done in moving a 10-µC charge from point A(1, 30°, 120°) to B(4, 90°, 60°). 4. **Work Done in Moving a Point Charge:** Find the amount of work done in moving a point charge Q = 3 µC from (4, π, 0) m to (2, π/2, 2) m in the field E = (10⁵/r)ar + 10⁵zaz (V/m). 5. **Uniform Surface Charge Density:** A uniform surface charge density of 20 nC/m² is present on the spherical surface r = 0.6 cm in free space. Find the absolute potential at P(r = 1 cm, θ = 25°, φ = 50°). 6. **Potential Difference:** A uniform surface charge density of 20 nC/m² is present on the spherical surface r = 0.6 cm in free space. Find the absolute potential at P(r = 1 cm, θ = 25°, ф = 50°), and VAB given points A(r = 2 cm, θ = 30°, ф = 60°) and B(r = 3 cm, θ = 45°, ф = 90°). 7. **Line Charges:** Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = −1, y = 2 in free space. If the potential at the origin is 100 V, find V at P(4, 1, 3). 8. **Point Charge and Line Charges:** A point charge Q = -8.0 nC at (-4, 2, 1) and a uniform line charge p₁ = 15 nC/m is at y = -2, z = -3. Given points A(3, 5, 1) and B(4, 0, 2), find the potential difference VAB if V = 4V at the (-1, 0, 0). 9. **Parallel Infinite Line Charges:** Two parallel infinite line charges, one having a charge density of 10 µC/m at x = 2 m and the other, with density of – 10 µC/m at x = -2 m are parallel along the z-axis. Find V at (1, 2). 10. **Electric Field and Potential Difference:** The electric field in a given region is E = (5/r)a, V/m for 0 ≤ r ≤ 2 m and E = 2.5ar for r > 2 m. Find the potential difference VAB for A(1, 0, 0) m and B(4, 0, 0) m. 11. **Electric Potential Function:** The electric potential in a region is given by V = 10x² + 20y² + 5z V. What is the electric field intensity? Can this potential function exist? 12. **Work Done in Transferring a Charge:** If V = 6x²yz, calculate the work done in transferring a 20 nC charge from (–2, 4, 0) to (4, 7, 3). # Chapter VI - Materials in Electric Fields ## Introduction - Materials can be classified as conductors, dielectrics, or semiconductors based on their electrical properties. - This chapter explores the behavior of static electric fields in conductors and dielectrics. - It provides a detailed discussion of the properties of conductors in electrostatic equilibrium, emphasizing their use as electrostatic shields. - The properties of dielectrics under the influence of electric fields are discussed, including polarization. - The chapter also presents Maxwell's equations and their applications in different media. - Boundaries between conductor-free space, conductor-dielectric, dielectric-free space, and dielectric-dielectric interfaces are explored. ## Classification of Materials - According to quantum theory, electrons in an atom can only occupy specific energy levels. - These energy levels are grouped into bands, each containing a number of discrete energy levels. - An electron can transition between energy levels through a quantum leap, requiring an exact amount of energy equal to the difference between the two levels. - This energy can be absorbed from an external source or released by the electron itself. ## Conductors Under Electrostatic Equilibrium - Conductors have a large number of free electrons due to their relatively small energy gaps. - The number of free electrons is extremely high in conductors with only one valence electron per atom. - Thermal agitation causes free electrons to move randomly, resulting in an absence of net drift in any direction. - A conductor in electrostatic equilibrium exhibits no net motion of charges. - An isolated conductor is always electrically neutral due to an equal number of protons and electrons. - The net charge density inside an isolated conductor is zero, and consequently there is no electric field inside the conductor. - When a conductor is charged, the excess charges move to the surface due to repulsive forces, setting the charge density and electric field inside the conductor to zero. - The surface of the conductor becomes equipotential, with a constant potential throughout. - The electric field just outside the conductor is perpendicular to the surface and proportional to the surface charge density. - The electric field outside the conductor is unaffected by the electric field inside the conductor. ## Electrostatic Shielding - Conductors can be used as electrostatic shields because the electric field inside a conductor is always zero. - The conductor effectively screens the external electric field, preventing it from reaching the interior. - For example, coaxial cables use a grounded outer conductor to shield the inner conductor from external fields. ## Illustrative Problem 6.1 Determine the electric field and electric potential anywhere inside, outside, and on the surface of a conducting sphere with a positive charge Q. **Solution:** - Inside the sphere, the electric field is zero due to the presence of free electrons moving to the surface. - Outside the sphere, we can use Gauss's law to find the electric field, which is the same as the field produced by a point charge Q at the center of the sphere: E = kQ/r² - The electric potential outside, V = kQ/r, is also the same as the potential due to a point charge Q at the center of the sphere. - The potential on the surface of the sphere, V = kQ/R (where R is the radius of the sphere), is the same as the potential at any point inside the conductor because the entire conductor is equipotential. ## Illustrative Problem 6.2 A uniformly charged nonconducting sphere of radius a is concentric with a conducting spherical shell of inner radius b and outer radius c. If the charge density on the solid sphere is pv, find the electric field everywhere, as well as the charge density on the inner and outer surfaces of the conducting shell. **Solution:** **(a)** For r < a: - Electric field E = (pvr / 3εο) **(b)** For a ≤ r < b: - Electric field E = (pva³ / 3εοr²) **(c)** For b ≤ r ≤ c: - Electric field E = 0 because the region is within the conductor. - The inner surface contains a charge density of psb = -pva³ / 3εοb² to balance the charge of the solid sphere. **(d)** For r > c: - Electric field E = (pva³ / 3εοr²) - The outer surface contains an equal amount of induced positive charges with a charge density of Psc = pva³ / 3εοc². ## Chapter Review **Part 1 - Multiple Choice:** 1. The main difference between conductors and insulators is the number of free electrons available for conduction. 2. A conductor is characterized by a constant electric potential throughout its volume. 3. In electrostatic equilibrium, the charge density and electric field inside a conductor are always zero. 4. The electric potential inside the conductor is always constant, not necessarily zero in electrostatic equilibrium. 5. The normal component of the electric field is continuous at the interface of a conductor and free-space if surface charge density exists. 6. Electric potential difference is defined as the change in potential energy per unit charge. 7. Electric field is a measure of the rate of change of the electric potential with respect to position. 8. Electric potential depends on the magnitude of electric charge. **Part 2 - Problems:** 1. **Electric Field and Charge Density:** The electric field at a point on the surface of a conductor in air is E = 20ax-30ay + 10az kV/m. Find the electric flux density and the charge density at that point. 2. **Potential in a Dielectric Material:** The potential in a material with a dielectric constant of 2.5 is defined by V = (50x + 20y² – 8xyz) V. Find: (A) electric flux density and (B) volume charge density. 3. **Transmission Line:** A transmission line consists of a cylindrical conductor with a charge density of 80 nC/m². The radius of the conductor is 15 mm, and it is surrounded by a plastic insulation (εr = 3.5) that is 10 mm thick. Find the electric field and flux density on the dielectric and outside the transmission line.

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