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 $



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