A) \[R_{2}^{2}/R_{1}^{2}\]
B) \[R_{1}^{2}/R_{2}^{2}\]
C) \[{{R}_{2}}/{{R}_{1}}\]
D) \[{{R}_{1}}/{{R}_{2}}\]
Correct Answer: C
Solution :
The sphere are joined by the wire, so their potential will be come equal. \[{{V}_{1}}={{V}_{2}}\] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{{{R}_{1}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{{{R}_{2}}}\] \[\therefore \] \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}\] \[\therefore \]Ratio of Electric fields \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{k\frac{{{q}_{1}}}{R_{1}^{2}}}{k\frac{{{q}_{2}}}{R_{2}^{2}}}\]where\[k=\frac{1}{4\pi {{\varepsilon }_{0}}}\] \[=\frac{{{q}_{1}}}{{{q}_{2}}}.{{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}\] \[=\frac{{{R}_{1}}}{{{R}_{2}}}.{{\left( \frac{{{R}_{1}}}{{{R}_{1}}} \right)}^{2}}\] \[=\frac{{{R}_{2}}}{{{R}_{1}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
