Physics, 26.02.2020 18:27 briannawoods13

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?

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Answer from: debbie4755

a) $1.64\cdot 10^{-4} J/m^3$

b) $0.88\cdot 10^{-4} J/m^3$

c) No

Explanation:

The energy density in a region with electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$ (1)

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

E is the magnitude of the electric field

Here the electric field between the two conducting shells is equal to the electric field given by the inner shell, so

$E=\frac{Q}{4\pi \epsilon_0 r^2}$ (2)

where Q is the charge stored in each plate of the capacitor, and r the distance from the centre.

We also know that the capacitance of a capacitor formed by two concentric spherical shells is

$C=\frac{Q}{V}=\frac{4\pi \epsilon_0}{\frac{1}{a}-\frac{1}{b}}$ (3)

where

V is the potential difference

a is the inner radius

b is the outer radius

From eq.(3) we find:

$Q=\frac{4\pi \epsilon_0 V}{\frac{1}{a}-\frac{1}{b}}$

And substituting into (2),

$E=\frac{1}{4\pi \epsilon_0 r^2}(\frac{4\pi \epsilon_0 V}{\frac{1}{a}-\frac{1}{b}})=\frac{V}{r^2}\frac{1}{\frac{1}{a}-\frac{1}{b}}$

And substituting into (1),

$u=\frac{1}{2}\epsilon_0 (\frac{V}{r^2}\frac{1}{\frac{1}{a}-\frac{1}{b}})^2$

Now we can apply this formula to find the energy density at different locations.

Here we want to find the energy density at

r = 12.6 cm = 0.126 cm

We have:

V = 120 V (potential difference)

a = 12.5 cm = 0.125 m (inner radius)

b = 14.8 cm = 0.148 m (outer radius)

Therefore, substituting into the equation, we find:

$u=\frac{1}{2}\epsilon_0 (\frac{120}{(0.126)^2}\frac{1}{\frac{1}{0.125}-\frac{1}{0.148}})^2=1.64\cdot 10^{-4} J/m^3$

Here we want to find the energy density at

r = 14.7 cm = 0.147 m

Therefore, substituting all the values of part a and this value of the radius, we find:

$u=\frac{1}{2}\epsilon_0 (\frac{120}{(0.147)^2}\frac{1}{\frac{1}{0.125}-\frac{1}{0.148}})^2=0.88\cdot 10^{-4} J/m^3$

For a parallel-plate capacitor the energy density is uniform in the region between the plates.

For a spherical capacitor, this is not true.

We can see this from the equation that gives the magnitude of the electric field:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

We see that the magnitude of the electric field depends on the distance from the centre, r. In particular, the electric field decreases as the distance increases.

We also know that the energy density depends quadratically on the magnitude of the electric field:

$u=\frac{1}{2}\epsilon_0 E^2$

Therefore, this means that the energy density depends on the distance r: so, it is not uniform.

Answer from: Quest

the largest known star in the universe is uy scuti, a hypergiant with a radius around 1,700 times larger than the sun.jul 26, 2018

explanation:

Answer from: Quest

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