question_answer

A particle of positive charge Q is fixed at point P. A second particle of mass m and negative charge - q moves at constant speed in a circle of radius \[{{r}_{1}}\], centered at P. The work W that must be done by an external agent on the second particle to increase the radius of the motion to \[{{r}_{2}}\], is given by

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                 \[W=F\times dr\]                 \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{r}_{1}}{{r}_{2}}}\times ({{r}_{2}}-{{r}_{1}})\]                 \[W=\frac{Qq}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\]