Question

Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential difference $ {{V}_{A}}-{{V}_{B}}={{V}_{B}}-{{V}_{C}}. $ If $ {{t}_{1}} $ and $ {{t}_{2}} $ be the distance between them. Then:

• A. $ {{t}_{1}}={{t}_{2}} $
• B. $ {{t}_{1}}>{{t}_{2}} $
• C. $ {{t}_{1}}<{{t}_{2}} $
• D. $ {{t}_{1}}\le {{t}_{2}} $

Answer 1

Answer: (C)

Potential difference between two equipotential surfaces A and B, $ {{V}_{A}}-{{V}_{B}}=kq\left( \frac{1}{{{r}_{A}}}-\frac{1}{{{r}_{B}}} \right) $ $ =kq\left( \frac{{{r}_{B}}-{{r}_{A}}}{{{r}_{A}}{{r}_{B}}} \right) $ $ =\frac{kq{{t}_{1}}}{{{r}_{A}}{{r}_{B}}} $ or $ {{t}_{1}}=\frac{({{V}_{A}}-{{V}_{B}}){{r}_{A}}{{r}_{B}}}{kq} $ or $ {{t}_{1}}\propto {{r}_{A}}{{r}_{B}} $ Similarly, $ {{t}_{2}}\propto {{r}_{B}}{{r}_{C}} $ Since $ {{r}_{A}}<{{r}_{B}}<{{r}_{C}}, $ therefore $ {{r}_{A}}{{r}_{B}}<{{r}_{B}}{{r}_{C}} $ $ \therefore $ $ {{t}_{1}}<{{t}_{2}} $