Electrostatics Question 777
Question: Four charges $ q _{1}=2\times {{10}^{-8}}C, $
$ q _{2}=-2\times {{10}^{-8}}C, $
$ q _{3}=-3\times {{10}^{-8}}C $ , $ q _{4}=6\times {{10}^{-8}}C $ are placed at four corners of a square of side $ \sqrt{2} $ m. What is the electric field at the center of the square?
Options:
A) 270 V
B) 300 V
C) Zero
D) 100 V
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Potential at the center O, $ V=k\frac{q}{r} $
$ V=k[ \frac{2\times {{10}^{-8}}}{1}+\frac{-2\times {{10}^{-8}}}{1}+\frac{-3\times {{10}^{-8}}}{1}+\frac{6\times {{10}^{-8}}}{1} ] $
$ V=k\times 3\times {{10}^{-8}}=9\times 10^{9}\times 3\times {{10}^{-8}}volt $
$ =27\times 10=270\ \text{V}$
BETA
