Electrostatics Question 331
Question: Charges of $ +\frac{10}{3}\times {{10}^{-9}}C $ are placed at each of the four corners of a square of side $ 8cm $ . The potential at the intersection of the diagonals is [BIT 1993]
Options:
A) $ 150\sqrt{2}volt $
B) $ 1500\sqrt{2}volt $
C) $ 900\sqrt{2}volt $
D) $ 900volt $
Show Answer
Answer:
Correct Answer: B
Solution:
Potential at the centre O,
$ V=4\times \frac{1}{4\pi {\varepsilon _{0}}}.\frac{Q}{a/\sqrt{2}} $ where $ Q=\frac{10}{3}\times {{10}^{-9}}C $
and $ a=8cm=8\times {{10}^{-2}}m $
So $ V=5\times 9\times 10^{9}\times \frac{\frac{10}{3}\times {{10}^{-9}}}{\frac{8\times {{10}^{-2}}}{\sqrt{2}}} $
$ =1500\sqrt{2}volt $
BETA
