question_answer

Two point charges +8q  and \[-2q\] are located at \[x=0\] and \[x=L\] respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero is                                          [AIEEE 2005]

A)            8 L

B)            4 L

C)            2 L

D)            \[\frac{L}{4}\]

Correct Answer: C

Solution :

             The net field will be zero at a point outside the charges and near the charge which is smaller in magnitude. Suppose E.F. is zero at P as shown. Hence at P; \[k.\frac{8q}{{{(L+l)}^{2}}}=\frac{k.(2q)}{{{l}^{2}}}\] Þ l = L. So distance of P from origin is L + L = 2L.