question_answer

A solid metal sphere of radius 50 cm carries a charge\[25\times {{10}^{-10}}C\]. The electrostatic potential at a distance of 20 cm from the centre will be:

A)  25V          

B)  15V

C)  35V          

D)  45V

Correct Answer: D

Solution :

 The potential at P is the ratio of the work done in bringing the test charge from infinity to the point P to the magnitude of test charge (q). Thus, the potential at P due to charge + q is given by \[V=\frac{W}{{{q}_{0}}}=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{q}{r}\]volt Given, \[r=50\,cm=50\times {{10}^{-2}}m\], \[q=25\times {{10}^{-10}}C\] \[\therefore \] \[V=\frac{9\times {{10}^{9}}\times 25\times {{10}^{-10}}}{50\times {{10}^{-2}}}\] V = 45 volt.