Chapter 4: Q32E (page 810)

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at r = 12.6 cm, just outside the inner sphere? (b) What is the energy density at r = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

Short Answer

Expert verified

(a)The energy density at $r = 12.6$ just outside the inner sphere is: $1.63 \times 10^{- 4} J . m^{-} 3$

(b) The energy density at $r = 12.6$ just outside the inner sphere is $8.8 \times 10^{- 5} J . m^{- 3}$

(c) No, the energy density is not uniformly distributed in the region between the plates except the near the edges of the plate.

Step by step solution

Calculating the density

Given,

$r_{b} = 14.8 c m r_{a} = 12.5 c m V = 120 V$

Now, to get density we need to find the charge for hollow spheres, and the potential difference between the surface is given by:

$V_{a b} = k Q (\frac{1}{r_{a}} - \frac{1}{r_{b}})$

Substituting the values, we get:

$Q = \frac{V_{a b}}{k (\frac{1}{r_{a}} - \frac{1}{r_{b}})} = \frac{120}{9 \times 10^{9} (\frac{1}{0.125} - \frac{1}{0.148})} = 10.72 \times 10^{- 9} C$

Now, the density is given by:

$u = \frac{Q^{2}}{32 π^{2} ε_{0} r^{4}}$

Substituting the values, we get:

$u = \frac{{(10.72 \times 10^{- 9})}^{2}}{32 π^{2} \times 8.85 \times 10^{- 12} \times {(0.126)}^{4}} = 1.63 \times 10^{- 4} J . m^{- 3}$

Therefore, the density is $1.63 \times 10^{- 4} J . m^{- 3}$

Energy distribution

We, know that density is given by:

$u = \frac{{(10.27 \times 10^{- 9})}^{2}}{32 π^{2} \times 8.85 \times 10^{- 12} \times {(0.147)}^{4}} = 8.8 \times 10^{- 5} J . m^{- 3}$

Therefore, we can conclude that the energy density is not uniformly distributed, but decreases while being close to the outer shell.

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Short Answer

Step by step solution

Calculating the density

Energy distribution

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