Electrostatics Question 389
Question: Four charges are placed on corners of a square having side of $ 5cm $ . If Q is one microcoulomb, then electric field intensity at centre will be [RPET 1999]
Options:
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
Show Answer
Answer:
Correct Answer: A
Solution:
Side a = 5x 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}{{{( \frac{a}{\sqrt{2}} )}^{2}}} $
Now field at O $ =\sqrt{E^{2}+E^{2}}=E\sqrt{2} $
$ =\frac{kq}{{{( \frac{a}{\sqrt{2}} )}^{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)
