Question

Two spherical conductors A and B of radii 1 mm and 2mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in the equilibrium condition the ratio of electric fields at surfaces of A and B is

• A. 4:01
• B. 1:02
• C. 2:01
• D. 1:04

Answer 1

The Correct Option is C

When the spherical conductors are connected by a conducting wire, charge is redistributed and the spheres attain a common potential $V$ . $\therefore $ Intensity $\text{E}_{\text{A}}=\frac{1}{4 \pi ?_{0}}\frac{\text{Q}_{\text{A}}}{\text{R}_{\text{A}}^{2}}$ or $\left(\text{E}\right)_{\text{A}}=\frac{1 \times \left(\text{C}\right)_{\text{A}} \text{V}}{4 \left(\pi ?\right)_{0} \text{R}_{\text{A}}^{2}}=\frac{\left(4 \left(\pi ?\right)_{0} \left(\text{R}\right)_{\text{A}}\right) \text{V}}{4 \left(\pi ?\right)_{0} \text{R}_{\text{A}}^{2}}=\frac{\text{V}}{\left(\text{R}\right)_{\text{A}}}$ Similarly $\text{E}_{\text{B}}=\frac{\text{V}}{\text{R}_{\text{B}}}$ $\therefore $ $\frac{\text{E}_{\text{A}}}{\text{E}_{\text{B}}}=\frac{\text{R}_{\text{B}}}{\text{R}_{\text{A}}}=\frac{2}{1}$ .