Question

A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞) on its surface. For this sphere, the equipotential surfaces with potentials 3Vo/2, 5Vo/4, 3V/4 and Vo/4 have radius R₁, R₂,R₃ and R4 respectively. Then

R₁ = 0 and R₂>(R₄-R₃)

R₁≠0 and (R₂-R₁)>(R₄-R₃)

R₁ = 0 and R₂<(R₄-R₃)

2R<R₄

Solution

R₁ = 0 and R₂<(R₄-R₃)

The potential at the surface of the charged sphere.

$straight V subscript 0 space equals space KQ over straight R straight V space equals space KQ over straight r comma straight r space greater or equal than space straight R equals space fraction numerator KQ over denominator 2 straight R cubed end fraction left parenthesis 3 straight R squared minus straight r squared right parenthesis semicolon space straight r less or equal than straight R straight V subscript centre space equals space Vc space equals space fraction numerator KQ over denominator 2 straight R cubed end fraction straight x 3 straight R squared space equals space fraction numerator 2 KQ over denominator 2 straight R end fraction space space fraction numerator 3 straight V subscript 0 over denominator 2 end fraction$
As potential decreases for outside points. Thus, according to the question, we can write
$straight V subscript straight R subscript 2 end subscript space equals fraction numerator 5 straight V subscript straight o over denominator 4 end fraction space equals fraction numerator KQ over denominator 2 straight R cubed end fraction left parenthesis 3 straight R squared minus straight R subscript 2 superscript 2 right parenthesis fraction numerator 5 straight V subscript 0 over denominator 4 end fraction space space equals space fraction numerator straight V subscript 0 over denominator 2 straight R squared end fraction left parenthesis 3 straight R squared minus straight R subscript 2 superscript 2 right parenthesis or space 5 over 2 space equals space 3 minus open parentheses straight R subscript 2 over straight R close parentheses squared open parentheses straight R subscript 2 over straight R close parentheses squared space equals space 3 minus 5 over 2 space equals space 1 half or space straight R subscript 2 space equals space fraction numerator straight R over denominator square root of 2 end fraction$
Similarly,
$straight V subscript straight R subscript 3 end subscript space equals space fraction numerator 3 straight V subscript 0 over denominator 4 end fraction KQ over straight R subscript 3 space equals space 3 over 4 space straight x space KQ over straight R or space straight R subscript 3 space equals space 4 over 3 straight R straight V subscript straight R subscript 4 end subscript space equals space KQ over straight R subscript 4 space equals space straight V subscript 0 over 4 KQ over straight R subscript 4 space equals space 1 fourth straight x KQ over straight R straight R subscript 4 space equals space 4 straight R$

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