Question

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.5cm , and the outer sphere has radius 16.5cm . A potential difference of 100V is applied to the capacitor.

What is the energy density at r= 11.6cm , just outside the inner sphere?

What is the energy density at r = 16.4cm , just inside the outer sphere?

Answer 1

Answer:

Explanation:

radius of inner sphere, a = 11.5 cm = 0.115 m

radius of outer sphere, b = 16.5 cm = 0.165 m

Voltage, V = 100 V

the capacitance of the spherical capacitor is given by

$C = \frac{4\pi \epsilon _{0}ab}{b-a}$

$C = \frac{0.115\times 0.165}{9\times 10^{9}\left ( 0.165-0.115 \right )}$

C = 4.22 x 10^-11 F

Charge, Q = c x V

Q = 4.22 x 10^-11 x 100

Q = 4.22 x 10^-9 C

(a) r = 11.6 cm = 0.116 m

The electric field at a distance r is given by

$E=\frac{KQ}{r^{2}}$

Where, K is the Coulombic constant

$E=\frac{9\times 10^{9}\times 4.22\times 10^{-9}}{0.116\times 0.116}$

E = 2820.3 N/C

The energy density is given by

$u = \frac{1}{2}\epsilon _{0}E^{2}$

$u = \frac{1}{2}\times 8.854\times 10^{-12}\times 2820.3^{2}$

u = 3.52 x 10^-5 J/m³

(b) r = 16.4 cm = 0.164 m

The electric field at a distance r is given by

$E=\frac{KQ}{r^{2}}$

Where, K is the Coulombic constant

$E=\frac{9\times 10^{9}\times 4.22\times 10^{-9}}{0.164\times 0.164}$

E = 1412.1 N/C

The energy density is given by

$u = \frac{1}{2}\epsilon _{0}E^{2}$

$u = \frac{1}{2}\times 8.854\times 10^{-12}\times 1412.1^{2}$

u = 8.83 x 10^-6 J/m³