College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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A spherical capacitor is formed from two conecentric, spherical, conducting shells separated by vacuum.  The inner sphere has a radius 15.0 cm and capacitance is 116 pF.  What is the radius of the outer sphere, and when the potential difference is 220.0 V, what is the magnitude of charge on each sphere?

Expert Solution
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Step 1: the radius of the outer sphere

Radius of inner sphere is r subscript a equals 15.0 space cm space equals space 0.15 space straight m

Capacitance is C equals 116 space pF equals 116 cross times 10 to the power of negative 12 end exponent space straight F

Outer radius = rb

Formula:

C equals 4 straight pi straight epsilon subscript 0 fraction numerator r subscript a r subscript b over denominator r subscript b minus r subscript a end fraction

Here, epsilon subscript 0 is free space permittivity having a value of 8.854 cross times 10 to the power of negative 12 end exponent space straight F divided by straight m

Let us rearrange the formula:

table row C equals cell 4 straight pi straight epsilon subscript 0 fraction numerator r subscript a r subscript b over denominator r subscript b minus r subscript a end fraction end cell row cell r subscript b C minus r subscript a C end cell equals cell 4 straight pi straight epsilon subscript 0 r subscript a r subscript b end cell row cell r subscript b end cell equals cell fraction numerator r subscript a C over denominator open parentheses C minus 4 straight pi straight epsilon subscript 0 r subscript a close parentheses end fraction end cell row cell r subscript b end cell equals cell fraction numerator 0.15 cross times 116 cross times 10 to the power of negative 12 end exponent over denominator open curly brackets open parentheses 116 cross times 10 to the power of negative 12 end exponent close parentheses minus 4 straight pi cross times 8.854 cross times 10 to the power of negative 12 end exponent cross times 0.15 close curly brackets end fraction space straight m end cell row cell r subscript b end cell equals cell 0.175 space straight m space equals space 17.5 space cm end cell end table


Answer: Outer radius is 17.5 cm

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