A) \[\frac{R\,_{2}^{2}}{R_{1}^{2}}\]
B) \[\frac{R\,_{1}^{2}}{R_{2}^{2}}\]
C) \[\frac{R{{\,}_{2}}}{{{R}_{1}}}\]
D) \[\frac{R{{\,}_{1}}}{{{R}_{2}}}\]
Correct Answer: C
Solution :
When two spheres arc joined charge flows till it equalizes. Hence electric potential is same. \[\therefore \] \[{{V}_{1}}={{V}_{2}}\] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{{{R}_{1}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{{{R}_{2}}}\] \[\frac{{{q}_{1}}}{{{R}_{1}}}=\frac{{{q}_{2}}}{{{R}_{2}}}\] ... (i) Ratio of electric fields\[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{R_{1}^{2}}}{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{R_{2}^{2}}}\] \[\Rightarrow \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{q}_{1}}}{{{q}_{2}}}{{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}\] ? (ii) or \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}}{{\left( \frac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}=\frac{{{R}_{2}}}{{{R}_{1}}}\] [From Eq.(i)]You need to login to perform this action.
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