Problem 1:
An infinite cylinder of radius R, with its symmetry axis oriented along the z-axis, is filled with uniform charge density r. Determine the electric field E(r), where r is the perpendicular distance from the z-axis.
Solution:
 | Concepts, principles, relations
that apply to the problem: Gauss’ law |
| Why do they apply? The problem has enough symmetry to find E(r) from Gauss’ law alone. |
| How do they apply? Use cylindrical coordinates (r, f, z). Symmetry dictates that E(r) = E(r). From Gauss’ law we have for a cylindrical Gaussian surface of radius R and length L 2prLE(r) = Qinside/e0. For r < R we have Qinside = pr2Lr. Therefore E(r) = rr/(2e0). For r > R we have Qinside = pR2Lr. Therefore E(r) = R2r/(2re0). |
| Details of the calculation: None |
Problem 2:
Determine the charge distribution that will give rise to the potential V(r) = kq exp(-mr)/r, with m a positive constants. Calculate the total charge in the distribution.
Solution:
| Concepts, principles, relations
that apply to the problem: Gauss' law, Ñ2V(r) = -r/e0, |
| Why do they apply? V(r) = V(r). Ñ2V(r) = -r/e0, except at r = 0, where the expression for Ñ2V(r) = is not defined. To find the charge at the origin we use Gauss’ law. |
| How do they apply? Ñ2V(r) = (1/r2)(¶/¶r)(r2¶V(r)/¶r) = m2 kq exp(-mr)/r = m2V(r) = -r/e0, except at r = 0, where the above expression for Ñ2V(r) = is not defined. To find the charge at the origin we use Gauss’ law. E = -ÑV(r) = (¶V(r)/¶r)(r/r). Er(r) = (kq/r2)exp(-mr) + kq m exp(-mr)/r. For a spherical surface of radius r we have 4pr2E(r) = Qinside/e0. As r à 0, Qinside/e0 = 4p kq. The charge distribution that gives rise to the potential V(r) = kq exp(-mr)/r therefore is r(r) = 4pe0 kqd(r) - e0m2 kq exp(-mr)/r. With k = 1/(4pe0 ) we have r(r) = qd(r) – (m2/4p)q exp(-mr)/r. The total charge in the distribution is given by Q = òall space r(r)dV = qòall space d(r)dV - (m2/4p)q 4p ò0¥ r2 exp(-mr)/r dr = q - m2q/m2 = 0. |
| Details of the calculation: None |
Problem 3:
A conductor at potential V = 0 has the shape of an infinite plane with a hemispherical bulge of radius R. A charge q is placed above the center of the bulge, a distance d from the plane (which means a distance d - R from the top of the bulge). What is the electrostatic force on the charge?
Solution:
| Concepts, principles, relations
that apply to the problem: The method of images |
| Why do they apply? The method of images is can be used to find the potential an field produced by a charge distribution outside a grounded conducting sphere and outside a grounded conducting sphere. Here we have a combination of these two geometries. |
| How do they apply? Let the z-axis be the symmetry axis. Placing image charges –q at z = - d, q' = -qR/d at z = R2/d and -q' = qR/d at z = -R2/d -on the z-axis makes the surface of the conductor an equipotential surface. The field everywhere outside the conductor is the same as that due to q and the image charges (uniqueness theorem). The force on q therefore is F = k(-q2/(4d2) + qq'/(d - R2/d)2 - qq'/(d + R2/d)2)k = -kq2(1/(4d2) + 4R3d3/(d4 - R4)2)k. The charge is pulled towards the conductor. |
| Details of the calculation: None |
Problem 4:
A vertical thin rod of length L carries a total charge Q uniformly distributed. Calculate the electric field along its axis at a distance z above its top end.
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? E(z) = kò-L0 ldz’/(z-z’)2 = klòz-Lz ldz’’/z’’2 = klò(1/(z-L) – 1/z) = kQ/((z-L)z). E = Ek. |
| Details of the calculation: None |
Problem 5:
A particle of mass m and charge e is suspended on a string of length L. At a distance d (d > L) under the point of suspension there is an infinite plane conductor. Ignore gravity. Compute the frequency of the pendulum if the amplitude is sufficiently small such that Hooke’s law is valid. (Ignore radiation losses).
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 or Lagrange's equations yield the equation of motion. |
| How do they apply? T = (1/2)mv2 = (1/2)mL2(dq/dt)2, U = -(1/(16pe0)(q2/(d-Lcosq)), Lagrangian = T - U. [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 work necessary to move the charge from infinity to its position) is U = -q2/(16pe0z).] mL2d2q/dt2 = -(q2/(16pe0))[Lsinq/(d-Lcosq)2] is the equation of motion. For small q we have cosq » 1, sinq » q. d2q/dt2 = -q2q/(16pe0mL(d-L)2). We have Hooke's law. The frequency for small oscillation is w. w2 = q2/(16pe0mL(d-L)2). |
| Details of the calculation: Using Newton's second law we have Fz = -q2/(16pe0z2) = -q2/(16pe0(d-Lcosq)2). The component in the direction of q is Fq = -q2sinq/(16pe0(d-Lcosq)2). This is the net force, since the component perpendicular to the direction of q is canceled by the tension in the string. F = ma, Fq = -q2sinq/(16pe0(d-Lcosq)2) = mLdq/dt. In the small angle approximation we retain only terms to first order in q, cosq » 1, sinq » q. mLdq/dt = -q2q/(16pe0(d-L)2), dq/dt = -q2q/(16pe0mL(d-L)2). |
Problem 6:
n identical spherical droplets of water, each charged to the same potential V, merge to form a larger droplet. What will be its potential?
Solution:
| Concepts, principles, relations
that apply to the problem: The potential of a spherically symmetric charge distribution of finite radius R. |
| Why do they apply? The potential at the surface of a spherically symmetric charge distribution is V = kQ/R. As we combine n drops, Q and R change. |
| How do they apply? We assume that initially the drops are very far apart, so that we can neglect their interactions. For one drop: V = kQ/R For n drops: R --> R' = n1/3R, Q --> Q' = nQ V' = kQ'/R' = knQ/(n1/3R) = n2/3V. |
| Details of the calculation: None |
Problem 7:
A flat, insulating disk of radius R lies in the xy plane and is centered
at the origin. It has uniform surface charge density
s.
(a) Find the electric potential on the axis of the disk.
(b) Find the limiting behavior of the potential when z >> R,
and z << R.
Solution:
| Concepts, principles, relations that apply to the
problem: The electric potential |
| Why do they apply? We are asked to find the potential due to a surface charge distribution in a situation with symmetry. |
| How do they apply? df(z) = 2prsdr/(r2+z2)1/2 is the potential on the axis due to a ring of radius r in Gaussian units. Integrating from r = 0 to r = R we find f(z) = 2ps{(R2+z2)1/2 - z}. Q = spR2, so f(z) = (2Q/R2){(R2+z2)1/2 - z}. If z >> R then f(z) » (2Q/R2){z(1 + R2/2z) - z} = Q/z. This is the potential of a point charge. From very far away the disk looks like a point. If z << R then f(z) = (2Q/R2){R - z} = constant - 2Qz/R2. The constant is arbitrary. Ez = -df(z)/dz = 2Q/R2 = 2ps. This is the field of an infinite plane with surface charge density s. |
| Details of the calculation: None |