Question

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 230.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.30 cm.
A) Calculate the capacitance.
B) Calculate the radius of the inner sphere.
C) Calculate the electric field just outside the surface of theinner sphere.

Answer 1

Answer:

(A). The capacitance is $15.2\times10^{-12}\ F$

(B). The radius of the inner sphere is 0.0327 m .

(C). The electric field just outside the surface of the inner sphere is $2.9458\times10^{4}\ V/m$

Explanation:

Given that,

Charge = 3.50 nC

Potential difference = 230.0 V

Radius = 4.30 cm

(a). We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{Q}{V}$

$C=\dfrac{3.50\times10^{-9}}{230.0}$

$C=15.2\times10^{-12}\ F$

(b). We need to calculate the radius of the inner sphere

Using formula of capacitance

$C=\dfrac{4\pi\epsilon_{0}r_{a}r_{b}}{r_{b}-r_{a}}$

Put the value into the formula

$15.2\times10^{-12}=4\pi\times8.85\times10^{-12}(\dfrac{r_{a}\times4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}})$

$\dfrac{r_{a}4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}}=\dfrac{15.2\times10^{-12}}{4\pi\times8.85\times10^{-12}}$

$\dfrac{r_{a}4.30\times10^{-2}}{4.30\times10^{-2}-r_{a}}=0.13667$

$r_{a}\times4.30\times10^{-2}=0.13667\times(4.30\times10^{-2}-r_{a})$

$r_{a}(4.30\times10^{-2}+0.13667)=0.13667\times4.30\times10^{-2}$

$r_{a}=\dfrac{0.13667\times4.30\times10^{-2}}{(4.30\times10^{-2}+0.13667)}$

$r_{a}=0.0327\ m$

(c). We need to calculate the electric field just outside the surface of the inner sphere

Using formula of electric filed

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{Q}{r_{a}^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-9}}{0.0327}$

$E=2.9458\times10^{4}\ V/m$

Hence, (A). The capacitance is $15.2\times10^{-12}\ F$

(B). The radius of the inner sphere is 0.0327 m .

(C). The electric field just outside the surface of the inner sphere is $2.9458\times10^{4}\ V/m$