question_answer

Two spheres of radius \[a\] and \[b\] respectively are charged and joined by a wire. The ratio of electric field of the spheres is                                                                                                            [CPMT 1999; JIPMER 2000; RPET 2000]

A)                    \[a/b\]

B)                                      \[b/a\]

C)                    \[{{a}^{2}}/{{b}^{2}}\]

D)                                      \[{{b}^{2}}/a\]

Correct Answer: B

Solution :

                     Joined by a wire means they are at the same potential. For same potential \[\frac{k{{Q}_{1}}}{{{a}_{1}}}=\frac{k{{Q}_{2}}}{{{a}_{2}}}\]Þ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{a}{b}\] Further, the electric field at the surface of the sphere having radius R and charge Q is \ \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{k{{Q}_{1}}/{{a}^{2}}}{k{{Q}_{2}}/{{b}_{2}}}=\frac{{{Q}_{1}}}{{{Q}_{2}}}\times \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}\]