Problem 1:
Consider 3 positive charges q at the vertices of an equilateral triangle. Each side has length l. Find the total force on each of the charges.
Solution:
 | Concepts, principles, relations that apply to the
problem: Coulomb's law,  ,
ke = 1/(4pe0) = 9 109
Nm2/C2, the principle of superposition |
| Why do they apply? We are asked to calculate the force exerted on a point charge by other charges. |
| How do they apply? Arrange the charges as shown in the figure.
The magnitude of F1 is F1 = F21cos30o
+ F31cos30o, and the direction of F1
is the y-direction. |
| Details of the calculation: None |
Problem 2:
Consider a line charge with line charge density l = Q/2a that extends along the x-axis from x = -a to x = +a. Find the electric field on the y-axis.
Solution:
| Concepts, principles, relations that apply to the
problem: The electric field due to a charge distribution, the principle of superposition |
| Why do they apply? We are asked to find the electric field due to a line charge distribution. |
| How do they apply? The field on the y-axis due to an infinitesimal element of charge ldx is given by  ,
 ,where  .On the y-axis the field due to the line charge therefore is given by  , from symmetry, and
 .Using  we have  .The electric field on the y-axis points in the positive y-direction for y > 0 and in the negative y-direction for y < 0. As y becomes very large, we can neglect a2 compared to y2 under the square root and then Ey = keQ/y2 µ 1/y2. From very far away, the line charge looks like a point charge. If, on the other hand, the line is very long, and we y << a, then we can neglect y2 compared to a2 under the square root and then Ey = keQ/ay µ 1/y. Near a very long line charge the electric field falls off as 1/distance, not as 1/distance2. |
| Details of the calculation: None |
Problem 3:
Inside a sphere of radius R and uniformly charged with the volume charge density r, there is a neutral spherical cavity of radius R1 with its center a distance a from the center of the charged sphere. If (R1+a) < R, find the electric field inside the cavity.
Solution:
| Concepts, principles, relations that apply to the
problem: Gauss’ law, the principle of superposition |
| Why do they apply? We can view the sphere with the cavity as a superposition of a sphere with radius R having a uniform charge density r and another sphere with radius R1 located in the cavity space having a uniform charge density –r. The field due to each of these spherical charge distributions can be found from Gauss’ law. |
| How do they apply? Let the center of the large sphere be located at the origin. E = E1 + E2. E1 = rr/3e0 for r < R. E2 = -r(r-a)/3e0 inside the cavity. Inside the cavity we therefore have E = rr/3e0 - rr/3e0 + ra/3e0 = ra/3e0 The field inside the cavity is constant and points into the direction of the vector a. |
| Details of the calculation: None |
Problem 4:
A charge distribution produces an electric field E = A(1-exp(-br))(r/r3) where A and b are constants. Find the net charge within the radius r = 1/b.
Solution:
| Concepts, principles, relations that apply to the
problem: Gauss’ law |
| Why do they apply? A radial field is produced by a spherically symmetric charge distribution. |
| How do they apply? E(r)4pr2 = Qinside/e0 in SI units. For r = 1/b we have A(1-e-1)4p = Qinside/e0. Qinside = A(1-e-1)4pe0. |
| Details of the calculation: None |
Problem 5:
Prove, carefully explaining your reasoning, that the solution of Ñ×E = r/e0, Ñ´E = 0, for E is unique. You may assume that the sources of E are bounded in space and that therefore the field vanishes at sufficiently large distances from the sources.
Solution:
| Concepts, principles, relations that apply to the
problem: The uniqueness theorem, Green's first identity  Green's first identity is derived starting with Green's theorem:  Set A = fÑy, with f,y arbitrary scalar fields. Ñ×A = Ñ×(fÑy) = fÑ2y + Ñf×Ñy  |
| Why do they apply? We are asked to prove the uniqueness theorem. |
| How do they apply? Ñ´E = 0 <==> E = -Ñf, E is the gradient of some scalar function. Ñ×E = r/e0 <==> Ñ2f = -r/e0. Consider the volume V bounded by surfaces at infinity. Assume we
know r(r) everywhere inside V and we know
that f = 0 at the boundaries. |
| Details of the calculation: None |
Problem 6:
(a) Show that the plates of an isolated parallel-plate capacitor
attract each other with a force F = (Q2/2e0A), where
±Q are the charges on the plates and A is
the area of each plate.
[HINT: Consider the work necessary to increase the plate
separation from x to x + dx.]
(b) How does your answer to part (a) change if, instead, one
maintains a constant potential difference between the plates of the capacitor?
Solution:
| Concepts, principles, relations that apply to the
problem: Work done by an external force Fext: W = Fext×Dr Energy stored in a capacitor: U = (1/2)(Q2/C) = (1/2)CV2. Capacitance of a parallel plate capacitor: C = e0A/x, x = plate separation, |
| Why do they apply? As we change the plate separation, the energy stored in the capacitor changes. DU = mechanical work done by external agent + work done by the battery. |
| How do they apply? (a) The battery is disconnected, the charge on the capacitor is constant. U = (1/2)(Q2x/e0A), DU = (1/2)(Q2Dx/e0A) = FextDx. Fext = dU/dx, F = -Fext = -dU/dx = -(Q2/2e0A) = -U/x, the force is attractive. (b) The battery is connected, the voltage across the plates is constant. The force must be the same, however, the argument arriving at the expression is different. U = (1/2)CV2, DU = (1/2)DCV2. DU = FextDx + DQV, dU/dx = Fext + VdQ/dx dU/dx = (1/2)V2dC/dx, Q = CV, dQ/dx = VdC/dx. Fext = -dU/dx, F = -Fext = dU/dx = (1/2)V2dC/dx = -U/x |
| Details of the calculation: None |
Problem 7:
A sphere of radius R carries a surface charge density
s = s0cosq.
(a) Derive the exact potential everywhere (both inside and
outside the sphere) assuming that it vanishes as r goes to infinity.
(b) Calculate the dipole moment of the charge distribution and
deduce the approximate form of the potential at points far from the sphere (r >> R).
(c) Compare (a) and (b). What can you conclude about the higher
multipoles?
Solution:
| Concepts, principles, relations that apply to the
problem: Boundary value problems,  is
the general solution to Laplace's equation for problems with azimuthal
symmetry. |
| Why do they apply?is the most general solution
inside and outside of the sphere, since r = 0
inside and outside of the sphere. To find the specific solution we
apply boundary conditions. |
| How do they apply? Assume f1(r,q) = ArP1(cosq) = Arcosq inside the sphere and f2(r,q) = (B/r2)P1(cosq) = (B/r2)cosq outside the sphere. The symmetry of the situation suggests this form of the solution. If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution. The potential vanishes as r goes to infinity. Boundary conditions: f is continuous across the boundary. f1(R,q) = f2(R,q). A = B/R3.  , df2/dr
- df1/dr |R = -s/e0.
-2(B/R3)cosq - Acosq
= -s0cosq/e0.-2A -A = -s0/e0. A = s0/3e0. f1(r,q) = (s0/3e0)rcosq inside the sphere and f2(r,q) = (s0R3/3e0r2)cosq outside the sphere. |
| Details of the calculation: (b) dipole moment: p = òs(r)r dA = k s0òcosq Rcosq 2pR2sinqdq = k s0(4p/3)R3. dipole potential: fdip = (1/4pe0)p×r/r3 = (s0R3/3e0r2)cosq. The potential outside the sphere is a pure dipole potential. The higher multipoles are zero. |
Problem 8:
Two parallel metallic plates (infinite extension) are kept at potentials f = 0 and are located at z = 0 and z = L. A point charge q is at z0. What is the electrostatic potential between the plates?
Solution:
| Concepts, principles, relations that apply to the
problem: Method of images: use infinitely many image charges |
| Why do they apply? This problem involves a point charge in front of conducting planes. |
| How do they apply?
We need image charges –q at z = 2nL – z0, n = integer. |
| Details of the calculation: None |
Problem 9:
A charge q is released from rest a distance d from an infinite conducting plane. How long will it take for the charge to strike the plane?
Solution:
| Concepts, principles, relations that apply to the
problem: The method of images |
| Why do they apply? This problem involves a point charge in front of a conducting plane. The method of images yields the potential energy of the point charge and the force on the point charge and Newton's second law yields the equation of motion. Energy conservation immediately gives us the first integral. |
| How do they apply? If the charge is a distance z from the conducting plane, it is a distance 2z from its image. The force on the charge is Fz = -q2/(16pe0z2), and its potential energy (the negative of the work necessary to remove the charge from its position to infinity) is U = -q2/(16pe0z). When a point charge moves from z = d to z = z', its potential energy changes by DU = U(z') - U(d). A point charge released from rest at z = d therefore has kinetic energy (1/2)mv2 = [q2/(16pe0)][1/z' - 1/d] at z'. v(z) = dz/dt = {[q2/(8mpe0)][1/z - 1/d]}1/2. |
| Details of the calculation: [Let z = dsin2q, dz = 2dsinqcosqdq,  .]T is the time it will take for the charge to strike the plane. |
Problem 10:
Centered at the origin of coordinates is an insulated, conducting sphere of radius a. A positive point charge q is located a distance d (d > a) from the center. What is the smallest positive charge, which must be on the sphere in order that the surface charge density will nowhere be negative?
Solution:
| Concepts, principles, relations that apply to the
problem: The method of images |
| Why do they apply? The method of images is the preferred method for finding the potential of a charge distribution outside a conducting sphere. |
| How do they apply? Assume that the point charge q is located on the z axis at z = d. Place an image charge q' = -aq/d on the z-axis at z' = a2/d. This will keep the sphere at zero potential. Assume the sphere must hold a total charge Q = q' + q'' in order that the surface charge density will nowhere be negative. Place q'' at the center of the sphere |
| Details of the calculation: s = e0Er is the surface charge density. Er is the radial component of the electric field on the surface.  The first term in the bracket is a negative number. Its magnitude is largest when the denominator is smallest.  We want smin = 0.  |
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