Problem 1:
A point charge Q is placed a distance L from the center of an isolated conducting spherical shell of radius R. Find the direction of the force acting on the charge in two cases, when the charge is inside and outside the sphere (R<L, R>L).
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 or inside a conducting spherical shell. |
| How do they apply? Place image charge Q’ = -QR/L at L’ = R2/L to make the spherical shell an equipotential surface with f = 0. Place Q’’ = -Q’ at the center of the sphere to keep the net charge on the spherical shell zero. Force on Q if L > R: F = kQ(Q’/L2 – Q’(L - R2/L)2) (r/r), i.e. F points towards the origin. Force on Q if L < R: F = kQ(Q’/L2 + Q’(R2/L - L)2) (r/r), i.e. F points away from the origin. |
| Details of the calculation: None |
Problem 2:
A solid conducting sphere of radius 2 cm has a charge of 8 microCoulomb. A
conducting spherical shell of inner radius 4 cm and outer radius 5 cm is
concentric with the solid sphere and has a charge of -4 microCoulomb.
(a) What is the magnitude and direction of the electric field at r = 1 cm?
(b) What is
the magnitude and direction of the electric field at r = 3 cm?
(c) What is
the magnitude and direction of the electric field at r = 4.5 cm?
(d) What is the magnitude and direction of the electric field at r = 7 cm?
Solution:
| Concepts, principles, relations that apply to the
problem: Gauss' law, electrostatic properties of conductors |
| Why do they apply? This is a problem involving charged conductors with spherical symmetry. |
| How do they apply? (a) The interior of a conductor is always field-free. E = 0 inside a conductor. The total net charge on a conductor always distributes itself over the surface of the conductor. (b) The spherical symmetry of the problem tells
us that the field must be radial and can only depend on the distance from
the center. The flux through a spherical surface of radius r must
therefore be equal to the field times the surface area 4pr2. (c) The interior of a conductor is always field-free. E = 0 inside a conductor. The total net charge on a conductor always distributes itself over the surface of the conductor. (d) At r = 7 cm Qinside is the total
charge on the sphere and the shell, (8-4)microC. |
| Details of the calculation: None |
Problem 3:
A cylindrical capacitor of length L has an inner cylinder
of radius a and an outer cylinder of radius b. A potential difference V0
is maintained between the cylinders.
(a) Find the magnitude of the electric field E(r) in the region a < r < b in
terms of V0, a, and b, assuming L >> a, b.
(b) Find the capacitance in terms of L, a, and b.
(c) Find the radius amin of the inner cylinder that minimizes E(r)
at the surface a = r for a given V0 and b.
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? (a) From Gauss’ law: E(r ) µ 1/r between the cylinders. Let the outer cylinder be grounded and V0 be positive. Then the direction of E is radially outward. E(r) = A/r, wit A positive. Then V0 = -òbaE(r)dr = Aòab(1/r)dr = A(lnb – lna) = Aln(b/a). A = V0/ln(b/a). E(r) = V0/(rln(b/a)). E(a) = V0/(aln(b/a)). (b) C = Q/V0 = e0E(a)2paL/V0 = 2pe0L/(ln(b/a)) = -2pe0L ln(a/b). (c) To find the minimum value of E(a), we set dE(a)/da|amin = 0. dE(a)/da|amin = 0 --> d [aln(b/a)]/da|amin = 0 --> ln(b/amin) = 1, b/e = amin. Then E(amin) = V0e/b = 2.718 V0/b. The extremum is aminimum. We can show this simply by setting a = b/2 and finding E(a) = 2.885 V0/b, or a = b/4 and finding E(a) = 2.885 V0/b. |
| Details of the calculation: None |
Problem 4:
Compare the strengths of the electric and the gravitational force between a proton and an electron.
Solution:
| Concepts, principles, relations that apply to the
problem: Coulomb's law, Newton's Law of gravitation |
| Why do they apply? Coulombs law gives the electrostatic force between two particles, Newton's Law of gravitation gives the gravitational force between two particles. |
| How do they apply? For the gravitational force we have
and for the electric force we have
The ratio of Fe/Fg therefore is given by |
| Details of the calculation: None |
Problem 5:
(a) A point charge q is located a distance d away from the center of
a grounded conducting sphere of radius a. Does the electrostatic force
between the charge and the sphere pull these two objects together or push them
apart. Why?
(b)
Now suppose the conducting sphere is not grounded and carries no net charge.
Does the electrostatic force pull the two objects together or push them apart?
Why?
Is the magnitude of the force greater or less than the force between the charge
and the grounded sphere? Why?
Solution:
| Concepts, principles, relations
that apply to the problem: The method of images |
| Why do they apply? Consider a charge q at z = d on the z - axis. Assume a grounded conducting sphere of radius R is centered on the origin. Then placing an image charge q' = -qR/d at d' = R2/d on the z-axis makes the sphere an equipotential with f = 0. Add q’’ at the center of the sphere to change the potential or the total charge on the sphere. |
| How do they apply? (a) The electrostatic force between the charge q and the grounded sphere has the same magnitude and direction as the electrostatic force between the charge q and its image charge q'. It is attractive. (b) To keep the net charge on the sphere zero we need two image charges, q' = -qR/d at d' = R2/d and q'' = qR/d at the center. The electrostatic force between the charge q and the sphere has the same magnitude and direction as the electrostatic force between the charge q and the image charge q' and q''. It is attractive since q' is closer to q than q''. The magnitude of the force is less than the force between the charge and the grounded sphere. |
| Details of the calculation: None |
Problem 6:
A thin dielectric of length L
meters lies along the x-axis with one end at the origin and the other at the
point (L, 0). It is charged with density λ = 2Qx/L2 Coulombs/meter
with a total charge of Q Coulombs.
(a) Calculate the field E(x) the potential V(x) at the point (x, 0) for
x > L.
(b) Evaluate these expressions in the limit x >> L.
Solution:
| Concepts, principles, relations that apply to the
problem: The electric field and potential due to a charge distribution |
| Why do they apply? We are asked to find the electric field and potential due to a line charge distribution. |
| How do they apply? (a) E(x) = kò0L ldx’/(x-x’)2 = (2kQ/L2)ò0L x’dx’/(x-x’)2 = (2kQ/L2) (ln(1-L/x) + L/(x-L)) V(x) = kò0L ldx’/(x-x’) = (2kQ/L2)ò0L x’dx’/(x-x’) = (2kQ/L2)(-L – x ln(1-L/x)) (b) For x >> L we have E(x) = (2kQ/L2)(-L/x – ½L2/x2 + L/x +L2/x2) = kQ/x2. For x >> L we have V(x) = (2kQ/L2)(-L – x(-L/x – ½L2/x2) = kQ/x. |
| Details of the calculation: None |
Problem 7:
A copper sphere of radius R contains a spherical cavity of
radius a. The center of the
cavity is a distance d from the
center of the sphere, and d and a are such that the cavity is
entirely within the sphere. There is a total charge Q on the sphere.
(a) Find the electric field within the cavity.
(b) Find the electric field outside the sphere.
(c)
Suppose a point charge q is placed at the center of the
cavity. What is the effect on the fields within the cavity and outside the
sphere?
Solution:
| Concepts, principles, relations that apply to the
problem: Properties of conductors, shielding |
| Why do they apply? A conductor shields the fields on both sides of a closed, conducting surface from each other. No matter how you rearrange the charges inside the cavity, the field outside the cavity will not change. No matter what you do to the external field, the field inside the cavity will not change. |
| How do they apply? (a) The field is zero inside the cavity (shielding). (b) The electric field outside the sphere is the same as that of a point charge Q located at the origin (where we placed the center of the large sphere) with no other charges present. (c) The electric field outside the sphere is the same as that of a point charge Q+q located at the origin (where we placed the center of the large sphere) with no other charges present. The electric field inside the cavity is the same as that of a point charge q at the center of the cavity with no other charges present. |
| Details of the calculation: None |
Problem 8:
A Wideroe linear accelerator consists of a series of cylinders connected to an alternating voltage. A charged particle is accelerated during its passage through the gaps between the cylinders. Inside the cylinders there is no acceleration. The gaps all have equal lengths d, but the lengths of the cylinders, which are much longer than d, are chosen so that particles arrive at the gaps at the right phase of the alternating voltage to be accelerated. Let there be N cylinders. Find the length of the nth cylinder for a proton to be accelerated to the energy of 100 MeV if the frequency of the accelerating voltage is 150MHz and if the acceleration across each gap is assumed to be constant.
Solution:
| Concepts, principles, relations that apply to the
problem: Kinematics |
| Why do they apply? The particle is accelerated in gaps of equal length. This is a non-relativistic problem, E << mc2. |
| How do they apply? Let d be the width of each gap and l = nd be the length of the region through which the particle has been accelerated when its velocity is v. Let a be the acceleration, a = qE/m = qV/(dm). v2 = 2al. Let l = nd, so vn2 = n2ad = n2qV/m. Here vn is the speed of the particle after the nth gap. The length of the nth section is Ln = vnT/2, where T is the period of the AC voltage. Ln2 = nqVT2/(2m). If there are N cylinders the NV = 100MV. T = 1/f = (1/1.5*108)s. Ln2 = 100*1.6*10-13Js2/[(1.5*108)2*2*1.67*10-27kg)(n/N) Ln = (n/N)1/2*46.1 cm |
| Details of the calculation: None |
Problem 9:
(a) Two capacitors have an equivalent capacitance of 8pF if connected in
parallel, and 2pF, if connected in series. Find C1 and C2.
(b) A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change when the plate separation is
doubled?
Solution:
| Concepts, principles, relations that apply to the
problem: Capacitors in series and parallel, energy stored in a capacitor |
| Why do they apply? We are given the equivalent capacitance if two capacitors in series and parallel and are asked to find C1 and C2. We are asked to determine how the energy stored in a parallel-plate capacitor changes when the plates are moved with respect to each other. |
| How do they apply? (a) C = C1 + C2 + C3 + C4 + …. for capacitors in parallel, (1/C1) + (1/C2) + …. = 1/C for capacitors in series. C1 + C2 = 8pF, (1/C1)+(1/C2) = 1/(2pF), C1 = C2 = 4pF. (b) U = (1/2)(Q2/C). Q is constant since the plates are disconnected from a battery. C = e0A/d, therefore U is proportional to d. If the plate separation is doubled, the stored energy is doubled. |
| Details of the calculation: None |
Problem 10:
A small, uniformly charged ball of mass m and charge q is suspended on the top of a spring with spring constant k under an infinitely large conductive surface. What is the period of small oscillations of the ball and the distance of the center of the oscillation from the surface, if this distance was L before the ball was charged?
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 electrostatic force on the point charge and Newton's second law or Lagrange's equations yield the equation of motion. |
| How do they apply? This is a 1-dimensional problem. Let the distance z denote the distance of the ball below the plane. [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).] Assume that an uncharged ball is in equilibrium a distance L below the plate. Let x be the displacement from this equilibrium position in the z-direction. T = (1/2)mv2 = (1/2)m(dx/dt)2, U = -q2/(16pe0(L+x)) + ½kx2 Lagrangian = T - U. md2x/dt2 = -(q2/(16pe0))[1/(L+x)2] – kx is the equation of motion. d2x/dt2 = -(q2/(16mpe0))[1/(L+x)2] – (k/m)x For small displacements x we have d2x/dt2 = -(q2/(16mL2pe0))[1 - 2x/L] – (k/m)x = -q2/(16mL2pe0)) – [(k/m) - q2/(8mL3pe0)]x d2x/dt2 = A – Bx x = x0 + x1exp(iwt) -w2x1exp(iwt) = A - Bx0 - Bx1exp(iwt) w2 = B = [(k/m) - q2/(8mL3pe0)] x0 = A/B = -(q2/(16mL2pe0))/[(k/m) - q2/(8mL3pe0)] The new equilibrium position is L + x0 = L - (q2/(16L2pe0))/[(k - q2/(8L3pe0)] |
| Details of the calculation: None |
Problem 11:
Find the value of Cx in the circuit shown if C1 = 3.3 nF, C2 = 6.8 nF and the voltage ratio V0/V1 is to be 5 (V1 is the voltage across capacitor C1).
| Concepts, principles, relations that apply to the
problem: Capacitors in series and parallel |
| Why do they apply? We are given a network of 3 capacitors connected to a battery. |
| How do they apply? Let C be the parallel equivalent of C1 and Cx. C = C1 + Cx. Let V2 be the voltage across capacitor C2. V0 = V1 + V2. The charge stored on C2 must be equal to the charge stored on the parallel combination C. Therefore V0 = Q/C + Q/C2, V1 = Q/C. V0/V1 = (C2 + C)/C2 = 5. C2 + C = 5C2, C = 4C2, Cx = 4C2 – C1. Cx = 27.2 nF – 3.3 nF = 23.9 nF. |
| Details of the calculation: None |
Problem 12:
A spherical charge distribution is given by
r = r0(1-r/a),
r
Solution:
| Concepts, principles, relations that apply to the
problem: Gauss' law |
| Why do they apply? The field due to a spherically symmetric charge distribution can be found from Gauss’ law. |
| How do they apply? (a) Q = 4pr0ò0a r2dr(1-r/a) = 4pr0[(a3/3)-(a3/4)] = 4pr0a3/12. (b) E = (1/(4pe0))Q/r2, radially outward for positive r0. E = (1/e0)r0a3/(12r2). V = (1/(4pe0))Q/r = (1/e0)r0a3/(12r). (c) E = (1/(4pe0)) Qinside/r2, radially outward for positive r0. Qinside = Q = 4pr0ò0r r’2dr’(1-r’/a) = 4pr0[(r3/3)-(r4/(4a))]. E = (1/e0)r0[(r/3)-(r2/(4a))]. V = V(a) + òraE(r)dr = V(a) + (r0/e0)òra[(r/3)-(r2/(4a))]dr = V(a) + (r0/e0)[(r2/6)-(r3/(12a))]ra = (r0/e0)[a2/12 + a2/6 - a2/12 - r2/6 + r3/(12a)] = (r0/e0)[a2/6 - r2/6 + r3/(12a)] (d) U = (e0/2)òall_space E×EdV U = (r02/2e0)ò0a[(r2/9) - (r3/(6a)) + (r4/(16a2))]4pr2dr + (r02a6/288e0)òa¥(1/r4)4pr2dr = (r02a5/e0)*0.0648. or U = (1/2)ò0ar(r)f(r)dV = (4p/2)òrar(r)f(r)r2dr = (2pr02/e0)ò0a(1-r/a)[a2/6 - r2/6 + r3/(12a)]r2dr = (r02a5/e0)*0.0648. |
| Details of the calculation: None |
Problem 13:
An electron is released from rest at one point in a uniform electric field and moves a distance of 10 cm in 10-7 s. What is the electric field strength and what is the voltage between the two points?
Solution:
| Concepts, principles, relations that apply to the
problem: Kinematics, the electrostatic force, F = qE. |
| Why do they apply? Kinematics lets us solve for a, F = ma = qE yields E. |
| How do they apply? Dx = ½at2, a = 2*Dx/t2 = (2*0.1/10-14)m/s2. qE = ma, E = ma/q = 9.1*10-31*2*1013/(1.6*10-19) N/C = 113.75 V/m |
| Details of the calculation: None |
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