question_answer

Four charges are placed on corners of a square as shown in figure having side of \[5\,cm\]. If Q is one microcoulomb, then electric field intensity at centre will be        [RPET 1999]

A)                    \[1.02\times {{10}^{7}}N/C\] upwards

B)                    \[2.04\times {{10}^{7}}N/C\] downwards

C)                    \[2.04\times {{10}^{7}}N/C\] upwards

D)                    \[1.02\times {{10}^{7}}N/C\] downwards

Correct Answer: A

Solution :

                             Side a = 5 ´ 10?2 m Half of the diagonal of the square \[r=\frac{a}{\sqrt{2}}\] Electric field at centre due to charge q \[E=\frac{kq}{{{\left( \frac{a}{\sqrt{2}} \right)}^{2}}}\] Now field at O \[=\sqrt{{{E}^{2}}+{{E}^{2}}}=E\sqrt{2}\]\[=\frac{kq}{{{\left( \frac{a}{\sqrt{2}} \right)}^{2}}}.\sqrt{2}\] \[=\frac{9\times {{10}^{9}}\times {{10}^{-6}}\times \sqrt{2}\times 2}{{{(5\times {{10}^{-2}})}^{2}}}\]\[=1.02\times {{10}^{7}}\,N/C\] (upward)