question_answer

The linear charge density of four quadrants or aring of radius a is as shown. Electric field at centre will be

A)                  \[\frac{2{{\eta }^{2}}}{g{{a}^{2}}}\]                        

B)  zero

C)                  \[\frac{{{\eta }^{2}}}{g{{a}^{2}}}\]                          

D)  \[\frac{{{\eta }^{2}}}{2g{{a}^{2}}}\]

Correct Answer: D

Solution :

                 Consider an elementary portion of the ring.  Electric field at the centre due this elementary portion is  \[{{v}_{P}}=v,\,\,{{v}_{Q}}=0\]                                 \[{{v}_{P}}={{v}_{Q}}=0\] If we consider upper portion of the ring x-components of electric field cancelled out. Net electric field will be due to addition of the y-components.                 \[{{v}_{P}}=0,{{v}_{Q}}=2v\]                 \[\eta \]               \[\lambda \]                                 \[\frac{2\eta }{\lambda }[1-{{e}^{-\lambda t}}]\] Similarly due to lower portion net electric field                 \[\frac{\eta }{2\lambda }[1-{{e}^{-\lambda t}}]\] \[\frac{\eta }{{{\lambda }^{2}}}[1-{{e}^{-{{\lambda }^{2}}t}}]\] Total electric field \[\frac{\eta }{\lambda }[1-{{e}^{-\lambda t}}]\]