Physics hw: what am I doing wrong? A solid conducting sphere (a) has radius 0.6 cm and charge -1.6 pC...?

Question

inside a conducting spherical shell (b). The shell has inner radius 1.7 cm, outer radius 4.2 cm, and a net charge 25.9 pC.
Q' = charge on inner surface of shell
Q'' = charge on outer surface of shell
Find:
a) Q''
b) magnitude of electric field of a point P midway between the sphere and the inner part of the shell
c) potential of P assuming potential at r = infinity is 0

This is what I've done:
a) Q'' = Q_a + Q_b = (-1.6 pC) - (25.9 pC) = -27.5 pC

b) E = Q_a/((4)(pi)(r^2)(e_0))
r = (r_a + r_b)/2 = ((0.6 cm) + (1.7 cm)) = 1.15 cm = 0.0115 m
E = (-1.6E-12 C)/((4)(pi)((0.0115 m)^2)(8.85E-12 C^2/Nm^2)) = -3.996E2 N/C

c) V = V_p - V_c
V_c = outside of shell
r_p = radius of P
r_c = outside radius of shell
V_p = V + V_c
V = -(k)(Q_a)(1/r_p)(1/r_c)
V_c = (k)((Q_a) + (Q_b))/(r_c)
V_p = -(k)(Q_a)(1/r_p - 1/r_c) + (k)((Q_a) + (Q_b))/(r_c) = (8.988E9 Nm^2/C^2)(-1.6E-12 C)(1/(0.0115 m) - 1/(0.042 m)) + (8.988E9 Nm^2/C^2)((-1.6E-12 C) + (25.9E-12 C))/(0.042 m) = 6.108 V
But the units don't add up

Answer 1

First, in part b, the dimension are in cm, but the formula using e0 = 8.85*10^-12 is in farad/meter. Convert the cm to m, and recalculate. EDIT: I see you did that, but the number I get is -108.3N/C. Look at the numbers: the 10^-12 in the charge cancels 10^12 in e0, leaving approx. -1.6/(4π*9*1.3*10^-4) which is about -1.6/0.015, or ~100.

In part a) you computed the charge on the outer part of the shell. By constructing a spherical gaussian volume that includes only the material of the shell, outer charge and external field, the external field can be determined to be Q''/(4*π*e0*r^2), where r is the distance from the center of the shell. The potential of the shell outer surface is then Q''/4*π*e0*r. Because the shell is conducting, that is also the potential of the inner surface. From part b) you know the formula for the internal field, Q_a/(4*πe0*r^2); the potential is then Q_a/(4*πe0*r^2) integrated over r from P to inner radius of the shell. This is Q_a/(4*πe0*r) [ r = P to r = r_bi] Add this to the shell potential to get the potential at P.

The potential of the shell is 5.861V, the potential difference to point P is -0.403V, and the sum is 5.458V

Answer 2

a) Is shell charge + 25.9 pC (25.9 pC POSITIVE)? If it is, then Q' is just + 1.6 pC, just the opposite of Q, the charge on the inner sphere; the remainder, Q'', is, clearly, the difference between these figures, + 24.3 pC.

b) For this, I get -108.74 V.
E = KQ/r² = 8.988 E9 × -1.6 E-12 / (1.15 E-2)² = -108.74 V.

c) A surefire method for solving this, is the superposition principle: compute potential at surface of shell, as if it were no charge on inner sphere; compute potential at point P, due to Q; add both. Bear in mind that charge in shell is 25.9 pC; in the presence of Q, this charge, which would be entirely at the surface of shell, splits in two parts, as described above; nevertheless, total charge is still 25.9 pC.

In applying superposition principle, you first find a partial potential, ignoring Q, then the other way around. Then,

V-shell = 8.988 E9 × 25.9 E-12 / 4.2 E-2 = 5.5426 V.
V-sphere = 8.988 E9 × -1.6 E-12 / 1.15 E-2 = -1.2505 V
V-P = V-shell + V-sphere = 5.5426 − 1.2505 = 4.2921 V

Answer 3

floor section S scales as r^2. the quantity V scales as ? r^2 dr = r^3/3, so as that V = S r/3. For V=r this reduces to a million = S/3, or S=3. *************** Remo, no implicit assumptions approximately cakes and appetizers. only symmetry, and Euclidean metric of course. greater carefully, use the Stokes theorem: ? r dS = ? ? r dV, r is the radius vector, ? r = D, D is the length of area. via symmetry this supplies |r| S = D V. If |r|=V then S=D. looks greater medical, however the essense is the comparable.