A spherical capacitor consists of two concentric spherical conductors. The inner one of radius R_1  maintained at potential V_1and the outer conductor of radius R_2   at potential  V_2 . The potential at a point P at a distance x from the centre (where R_2> x> R_1 ) is

  • Option 1)

    \frac{V_1-V_2}{R_2-R_1}(x-R_1)

  • Option 2)

    \frac{(V_1R_1(R_2-x)+V_2(x-R_1))}{(R_2-R_1)x}

  • Option 3)

    V_1+\frac{V_2x}{(R_2-R_1)}

  • Option 4)

    \frac{(V_1+V_2)}{R_1+R_2}x

 

Answers (1)

As we have learned

Potential Difference -

V= \frac{Q}{4\pi \epsilon _{0}a}-\frac{Q}{4\pi \epsilon _{0}b}

 

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Let Q_1  and Q_2  be the charges on the inner and the outer sphere respectively. Now  is the total potential on the sphere of radius R1,

So, V_1 = \frac{Q_1}{R_1} + \frac{Q_2}{R_2}........(1)             

and V_2  is the total potential on the surface of sphere of radius R_2 ,

So, V_2 = \frac{Q_2}{R_2} + \frac{Q_1}{R_1}........(2)           

If V be the potential at point P which lies at a distance x from the common centre then

      V = \frac{Q_1}{x} = \frac{Q_1}{R_1} + V_1- \frac{Q_1}{R_1}= Q_1 \left ( \frac{1}{x} - \frac{1}{R_1} \right )+ V_1= \frac{Q_1(R_1-x)}{xR_1}+ V_1     ……..(iii)

                        Substracting (ii) from (i)

V_1- V_2 = \frac{Q_1}{R_1}-\frac{Q_2}{R_2}\Rightarrow (V_1-V_2) R_1R_2= R_2Q_1\Rightarrow Q_1 = \frac{(V_1-V_2)R_1R_2}{R_2-R_1}


Option 1)

\frac{V_1-V_2}{R_2-R_1}(x-R_1)

Option 2)

\frac{(V_1R_1(R_2-x)+V_2(x-R_1))}{(R_2-R_1)x}

Option 3)

V_1+\frac{V_2x}{(R_2-R_1)}

Option 4)

\frac{(V_1+V_2)}{R_1+R_2}x

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