A) \[W=\frac{Qq}{8\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\]
B) \[W=\frac{Qq}{8\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\]
C) \[W=\frac{Qq}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\]
D) \[W=\frac{Qq}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\]
Correct Answer: D
Solution :
Force at P due to q when the radius of circle is \[{{F}_{1}}=\frac{Qq}{4\pi {{\varepsilon }_{0}}r_{1}^{2}}\] ?. (i) Force at P due to q when the radius of circle is \[{{F}_{2}}=\frac{1}{4\pi \,{{\varepsilon }_{0}}}\frac{Qq}{r_{1}^{2}}\] .... (i) Increase in radius \[({{r}_{2}}-{{r}_{1}})\] (distance) Resultant force \[=\sqrt{{{F}_{1}}\times {{F}_{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{r}_{1}}r_{1}^{2}}\] \[\therefore \] Work done \[W=F\times dr\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{r}_{1}}{{r}_{2}}}\times ({{r}_{2}}-{{r}_{1}})\] \[W=\frac{Qq}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\]You need to login to perform this action.
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