Question

Two spheres of radius $R_1$ and $R_2$ respectively are charged and joined by a wire. The ratio of electric field of the spheres is:

• A. $\frac{R{}_2^2}{R{}_1^2}$
• B. $\frac{R{}_1^2}{R{}_2^2}$
• C. $\frac{R{}_2}{R{}_1}$
• D. $\frac{R{}_1}{R{}_2}$

Answer 1

Answer: (C)

When two spheres are joined by the wire by doing so we observe that, they are at same potential and for same potential $\frac{K{q}_1}{R_1}=\frac{K{q}_2}{R_2}$ or $\frac{q_1}{q_2}=\frac{R_1}{R_2}$ Again the electric field at the surface of the sphere having radius R and charge q is given by $E=\frac{Kq}{R^2}$ Hence $\frac{E_1}{E_2}=\frac{\frac{K{q}_1}{R_1^2}}{\left(\frac{K{q}_2^2}{R_2^2}\right)}=\frac{q_1}{q_2}\times \frac{R_2^2}{R_1^2}$ $=\frac{R_1}{R_2}\times \frac{R_2^2}{R_1^2}=\frac{R_2}{R_1}.$