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A spherical capacitor contains a solid spherical conductor of radius 0.5 mm with a charge of 7.4...

Question:

A spherical capacitor contains a solid spherical conductor of radius 0.5 mm with a charge of 7.4 micro coulombs, surrounded by a dielectric material with er = 1.8 out to a radius of 1.2 mm, then an outer spherical non-conducting shell, with variable charge per unit volume p = 5r, with outer radius 2.0 mm. Determine the electric field everywhere. (Remember that in a linear dielectric material you can work out the equations as though they are in a vacuum and then replace e0 by e0er.)

Electric Field:

The electric field at different points due to a conductor is obtained from the Coulomb constant, magnitude of the charge and the distance of the point from the charge. If a dielectric medium surrounds the charge, the effective permeability of the point needs to be calculated.

Answer and Explanation:

Given data:

  • The radius of the solid spherical conductor is {eq}{r_s} = 0.5\;{\rm{mm}}{/eq}.
  • The charge of the spherical conductor is {eq}Q = 7.4\;\mu {\rm{C}}{/eq}.
  • The permeability of dielectric material is {eq}{\varepsilon _r} = 1.8{/eq}.
  • The variable charge per unit volume of nonconducting shell is {eq}\rho = 5 \times r{/eq}.
  • Outer radius of the shell is {eq}{r_o} = 2\;{\rm{mm}}{/eq}.


There are three region in the conductor, first region is from 0 to 0.5 mm, second region is from 0.5 to 1.2 mm and the third region is from 1.2 mm to 2 mm.

Expression for the electric field in the first region is,

{eq}{E_1} = \dfrac{{k \times Q}}{{{r^2}}}{/eq}

Since the point is inside the conductor the magnitude of electric field is zero.

Therefore,

{eq}{E_1} = 0\;{\rm{N/C}}{/eq}

Expression for the electric field in the second region is,

{eq}{E_2} = \dfrac{{{k_r} \times Q}}{{{r^2}}}{/eq}

Substitute the values in the expression.

{eq}\begin{align*} {E_2} &= \dfrac{Q}{{4\pi \times {\varepsilon _0} \times {\varepsilon _r} \times {r^2}}}\\ &= \dfrac{{7.4 \times {{10}^{ - 6}}\;}}{{4\pi \times 8.85 \times {{10}^{ - 12}} \times 1.8 \times {{\left( {1.2 \times {{10}^{ - 3}}} \right)}^2}}}\\ &= 2.88 \times {10^{ - 16}}\;{\rm{N/C}} \end{align*}{/eq}

Thus the electric field in the second region is {eq}2.88 \times {10^{ - 16}}\;{\rm{N/C}}{/eq}.

Expression for the electric field in the third region is,

{eq}{E_3} = \dfrac{{\left( {Q + 5\pi } \right) \times \left( {r_0^4 - {a^4}} \right)}}{{{\varepsilon _0} \times 4\pi \times r_o^2}}{/eq}

Where,

a is the initial point of the nonconducting shell, {eq}a = 1.2\;{\rm{mm}}{/eq}

Substitute the values in the above expression.

{eq}\begin{align*} {E_3} &= \dfrac{{\left( {Q + 5\pi } \right) \times \left( {r_0^4 - {a^4}} \right)}}{{{\varepsilon _0} \times 4\pi \times r_o^2}}\\ &= \dfrac{{\left( {7.4 \times {{10}^{ - 6}} + 5\pi } \right) \times \left( {{{\left( {2 \times {{10}^{ - 3}}} \right)}^4} - {{\left( {1.2 \times {{10}^{ - 3}}} \right)}^4}} \right)}}{{8.85 \times {{10}^{ - 12}} \times 4\pi \times {{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}}\\ &= 4.91 \times {10^5}\;{\rm{N/C}} \end{align*}{/eq}

Thus the electric field at the third region is {eq}4.91 \times {10^5}\;{\rm{N/C}}{/eq}.


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