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) [tex]C=16.7pF[/tex]
(b) [tex]r_{a} =3.749cm[/tex]
(c) [tex]E=2.24*10^{4} N/C[/tex]
Explanation:
Given data
[tex]Q=3.50nC\\V=210V\\r_{b}=5.0cm[/tex]
For part (a)
The Capacitance given by:
[tex]C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF[/tex]
For part (b)
The Capacitance of coordinates is given as
[tex]C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm[/tex]
For part (c)
The electric field according to Gauss Law is given by:
[tex]EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C[/tex]
