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 :
2 : 1
Solution :
After connection, $${V_1} = {V_2}$$
$$\eqalign{ & \Rightarrow K\frac{{{Q_1}}}{{{r_1}}} = K\frac{{{Q_2}}}{{{r_2}}} \cr & \Rightarrow \frac{{{Q_1}}}{{{r_1}}} = \frac{{{Q_2}}}{{{r_2}}} \cr} $$
The ratio of electric fields
$$\eqalign{ & \frac{{{E_1}}}{{{E_2}}} = \frac{{K\frac{{{Q_1}}}{{r_1^2}}}}{{K\frac{{{Q_2}}}{{r_2^2}}}} = \frac{{{Q_1}}}{{r_1^2}} \times \frac{{r_2^2}}{{{Q_2}}} \cr & \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{r_1} \times r_2^2}}{{r_1^2 \times {r_2}}} \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{r_2}}}{{{r_1}}} = \frac{2}{1} \cr} $$
Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.
After connection, $${V_1} = {V_2}$$
$$\eqalign{ & \Rightarrow K\frac{{{Q_1}}}{{{r_1}}} = K\frac{{{Q_2}}}{{{r_2}}} \cr & \Rightarrow \frac{{{Q_1}}}{{{r_1}}} = \frac{{{Q_2}}}{{{r_2}}} \cr} $$
The ratio of electric fields
$$\eqalign{ & \frac{{{E_1}}}{{{E_2}}} = \frac{{K\frac{{{Q_1}}}{{r_1^2}}}}{{K\frac{{{Q_2}}}{{r_2^2}}}} = \frac{{{Q_1}}}{{r_1^2}} \times \frac{{r_2^2}}{{{Q_2}}} \cr & \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{r_1} \times r_2^2}}{{r_1^2 \times {r_2}}} \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{r_2}}}{{{r_1}}} = \frac{2}{1} \cr} $$
Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.
