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)



NCERT Chapter Video Solution

Dual Pane