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Question: A spherical capacitor consists of two concentric spherical conductors. The inner one of radius \(R _...

A spherical capacitor consists of two concentric spherical conductors. The inner one of radius R2R _ { 2 } at potentialV2V _ { 2 }. The potential at a point P at a distance x from the centre (where R2>x>R1R _ { 2 } > x > R _ { 1 }) is

A

V1−V2R2−R1(x−R1)\frac { V _ { 1 } - V _ { 2 } } { R _ { 2 } - R _ { 1 } } \left( x - R _ { 1 } \right)

B

V1R1(R2−x)+V2R2(x−R1)(R2−R1)x\frac { V _ { 1 } R _ { 1 } \left( R _ { 2 } - x \right) + V _ { 2 } R _ { 2 } \left( x - R _ { 1 } \right) } { \left( R _ { 2 } - R _ { 1 } \right) x }

C

V1+V2x(R2−R1)V _ { 1 } + \frac { V _ { 2 } x } { \left( R _ { 2 } - R _ { 1 } \right) }

D

(V1+V2)(R1+R2)x\frac { \left( V _ { 1 } + V _ { 2 } \right) } { \left( R _ { 1 } + R _ { 2 } \right) } x

Answer

V1R1(R2−x)+V2R2(x−R1)(R2−R1)x\frac { V _ { 1 } R _ { 1 } \left( R _ { 2 } - x \right) + V _ { 2 } R _ { 2 } \left( x - R _ { 1 } \right) } { \left( R _ { 2 } - R _ { 1 } \right) x }

Explanation

Solution

Let Q1Q _ { 1 } and V1V _ { 1 } is the total potential on the sphere of radius R1,

So, V1=Q1R1+Q2R2V _ { 1 } = \frac { Q _ { 1 } } { R _ { 1 } } + \frac { Q _ { 2 } } { R _ { 2 } } …….. (i) and V2V _ { 2 } is the total potential on the surface of sphere of radius R2R _ { 2 },

So, V2=Q2R2+Q1R2V _ { 2 } = \frac { Q _ { 2 } } { R _ { 2 } } + \frac { Q _ { 1 } } { R _ { 2 } } …….. (ii) If V be the potential at point P which lies at a distance x from the common centre then

=Q1(1x−1R1)+V1=Q1(R1−x)xR1+V1= Q _ { 1 } \left( \frac { 1 } { x } - \frac { 1 } { R _ { 1 } } \right) + V _ { 1 } = \frac { Q _ { 1 } \left( R _ { 1 } - x \right) } { x R _ { 1 } } + V _ { 1 } ……..(iii)

Substracting (ii) from (i)

V1−V2=Q1R1−Q2R2V _ { 1 } - V _ { 2 } = \frac { Q _ { 1 } } { R _ { 1 } } - \frac { Q _ { 2 } } { R _ { 2 } } ⇒ (V1−V2)R1R2=R2Q1−R1Q1\left( V _ { 1 } - V _ { 2 } \right) R _ { 1 } R _ { 2 } = R _ { 2 } Q _ { 1 } - R _ { 1 } Q _ { 1 }

⇒ Q1=(V1−V2)R1R2R2−R1Q _ { 1 } = \frac { \left( V _ { 1 } - V _ { 2 } \right) R _ { 1 } R _ { 2 } } { R _ { 2 } - R _ { 1 } }

Now substituting it in equation (iii), we have

V=(R1−x)(V1−V2)R1R2xR1(R2−R1)+V1V = \frac { \left( R _ { 1 } - x \right) \left( V _ { 1 } - V _ { 2 } \right) R _ { 1 } R _ { 2 } } { x R _ { 1 } \left( R _ { 2 } - R _ { 1 } \right) } + V _ { 1 }

⇒ V=V1R1(R2−x)+V2R2(x−R1)x(R2−R1)V = \frac { V _ { 1 } R _ { 1 } \left( R _ { 2 } - x \right) + V _ { 2 } R _ { 2 } \left( x - R _ { 1 } \right) } { x \left( R _ { 2 } - R _ { 1 } \right) }