Aspherical capacitor contains a charge of 3.10 nc when connected to a potential difference of 250 v. if its plates are separated by vacuum and the inner radius of the outer shell is 4.00 cma)calculate the capacitance: found to be 1.24*10^-11 fb)find the radius of the inner spere: found to be .0294mc)calculate the electric field just outside the surface of the inner sphere. t

(a)

The general formula for the capacitance of a capacitor is:

where

Q is the charge stored on the capacitor

V is the potential difference across the capacitor

In this problem, we have

is the charge stored

V = 250 V is the potential difference

Substituting, we find

(b) 0.0294 m

The capacitance of a spherical capacitor is given by

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

So we can re-arrange the equation to find a, the radius of the inner sphere:

(c)

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:

where

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:

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.