(a) [tex]1.24\cdot 10^{-11} F[/tex]
The general formula for the capacitance of a capacitor is:
[tex]C=\frac{Q}{V}[/tex]
where
Q is the charge stored on the capacitor
V is the potential difference across the capacitor
In this problem, we have
[tex]Q=3.10 nC = 3.10\cdot 10^{-9} C[/tex] is the charge stored
V = 250 V is the potential difference
Substituting, we find
[tex]C=\frac{3.10\cdot 10^{-9}C}{250 V}=1.24\cdot 10^{-11} F[/tex]
(b) 0.0294 m
The capacitance of a spherical capacitor is given by
[tex]C=\frac{4\pi \epsilon_0}{\frac{1}{a}-\frac{1}{b}}[/tex]
where
a is the radius of the inner shell
b is the radius of the outer shell
Here we have
b = 4.00 cm = 0.04 m
and the capacitance is
[tex]C=1.24\cdot 10^{-11} F[/tex]
So we can re-arrange the equation to find a, the radius of the inner sphere:
[tex]a=(\frac{4\pi \epsilon_0}{C}+\frac{1}{b})^{-1} =(\frac{4\pi (8.85\cdot 10^{-12}F/m)}{1.24\cdot 10^{-11}F}+\frac{1}{0.04 m})^{-1}=0.0294 m[/tex]
(c) [tex]3.225 \cdot 10^4 V/m[/tex]
The electric field just outside the surface of the inner sphere with charge Q is equal to the electric field produced by a single point charge Q at a distance of r = a:
[tex]E=\frac{Q}{4\pi \epsilon_0 a^2}[/tex]
where
[tex]Q=3.10 \cdot 10^{-9} C[/tex] is the charge on the sphere
a = 0.0294 m is the radius of the inner sphere
Substituting the data into the formula, we find:
[tex]E=\frac{3.10\cdot 10^{-9}C}{4\pi (8.85\cdot 10^{-12}F/m) (0.0294 m)^2}=3.225 \cdot 10^4 V/m[/tex]
