Question

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 10.5 cm, and the outer sphere has radius 16.5 cm. A potential difference of 150 V is applied to the capacitor. (a) What is the energy density at r = 10.6 cm, just outside the inner sphere

Answer 1

Answer:

$U = 2.91 *10^{-5} J$

Explanation:

energy density can be obtained as

$U = \frac{1}{2}\epsilon_o E^{2}$

Where,

E is electric field $= \frac{kQ}{R^{2}}$

K COLOUMB CONSTANT =8.99*10^{9} N -m2 /C2

Q is charge = CV

C is capacitance = $4\pi \epsilon_o \frac{r_1 r_2}{r_2 -r_1}$

                             $=4\pi *8.85*10^{-12} [\frac{10.5*16.5}{16.5 -10.5}]$

                           $= 3.21*10^{-9} F$

$Q = 3.21*10^{-9} *150 = 4.81*10^{-7] C$

for r  = 10.6 cm

$E = \frac{8.99*10^{9}*3.21*10^{-9}}{0.106^{2}}$

E = 2568.34 N/C

$U = \frac{1}{2}\epsilon_oE^{2}$

$U = \frac{1}{2}*8.85*10^{-12} *2568.34 ^2$

$U = 2.91 *10^{-5} J$