Test Index

Electrostatics

Section: Physics

Question : 45 of 59

Marks: +1, -0

R_A

R_B

R_A

Solution:

We know that the electric field is zero inside a conductor if it is hollow or solid. That is,

E_A^{\text{inside}}

= 0

On connecting them by a wire, both acquire a common potential, namely,

V_0

\frac{Q+Q}{C_1+C_2}

\frac{2Q}{4\pi\varepsilon_0 R_A+R_B)}

Therefore,

Q_A

C_1V_0

(4\pi\varepsilon_0 R_A)V_0

\frac{2Q R_A}{R_A+R_B}

Q_B

C_2V_0

(4\pi\varepsilon_0 R_B)V_0

\frac{2Q R_B}{R_A+R_B}

Since

R_A

R_B

, we get

Q_A

Q_B

. Now,

\frac{\sigma_A}{\sigma_B}

\frac{Q_A}{R_A^2} \times \frac{R_B^2}{Q_B}

\frac{R_A}{R_B} \times \frac{R_B^2}{R_A^2}

\frac{R_B}{R_A}

The electric field on the surface is

\frac{\sigma}{\varepsilon_0}

. Since

\sigma_A

\sigma_B

. we get

E_A^{\text{on surface}}

E_B^{\text{on surface}}

Go to Question:

Prev Question

Next Question