Question

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

• A. $4 : 1$
• B. $1 : 2$
• C. $2 : 1$
• D. $1 : 4$

Answer 1

Answer: (C)

When the two conducting spheres are connected by a conducting wire, charge will flow from one sphere (having higher potential) to other (having lower potential) till both acquire the same potential.
After connection, $V _{1}= V _{2}$ $\Rightarrow K \frac{ Q _{1}}{ r _{1}}= K \frac{ Q _{2}}{ r _{2}}$
$\Rightarrow \frac{ Q _{1}}{ r _{1}}=\frac{ Q _{2}}{ r _{2}}$
The ratio of electric fields $\frac{ E _{1}}{ E _{2}}=\frac{ K \frac{ Q _{1}}{ r _{1}^{2}}}{ K \frac{ Q _{2}}{ r _{2}^{2}}}$
$\Rightarrow \frac{ E _{1}}{ E _{2}}=\frac{ Q _{1}}{ r _{1}^{2}} \times \frac{ r _{2}^{2}}{ Q _{2}}$
$\Rightarrow \frac{ E _{1}}{ E _{2}}=\frac{ r _{1} \times r _{2}^{2}}{ r _{1}^{2} \times r _{2}} \Rightarrow \frac{ E _{1}}{ E _{2}}=\frac{ r _{2}}{ r _{1}}=\frac{ 2 }{ 1 }$