Question

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A spherical capacitor.?

Answer 1

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a)
`C=16.7pF`

(b)
`r_(a) =3.749cm`

(c)
`E=2.24*10^(4) N/C`

Step-by-step explanation:

Given data

`Q=3.50nC\\V=210V\\r_(b)=5.0cm`

For part (a)

The Capacitance given by:

`C=(Q)/(V)\\ C=(3.50*10^(-9) C)/(210V)\\C=1.6666*10^(-11)F\\or\\C=16.7pF`

For part (b)

The Capacitance of coordinates is given as

`C=(4\pi e)/((1)/(r_(a) )-(1)/(r_(b) ) )\\ So\\{(1)/(r_(a) )-(1)/(r_(b) ) }=(4\pi *8.85*10^(-12) )/(1.666*10^(-11))=6.672m^(-1) \\ (1)/(r_(a) )=6.672+(1 /0.05)\\(1)/(r_(a) )=26.672\\r_(a) =1/26.672\\r_(a) =0.0375m\\r_(a) =3.749cm`

For part (c)

The electric field according to Gauss Law is given by:

`EA=(Q)/(e)\\ E=(Q)/(4\pi er_(a)^(2) )=(kQ)/(r_(a)^(2))\\ E=(9*10^(9)*3.50*10^(-9) )/((0.0375m)^(2) )\\ E=2.24*10^(4) N/C`