Electrostatics Question 368

Question: Two positive charges of 20 $ coulomb $ and $ Q\ coulomb $ are situated at a distance of $ 60cm $ . The neutral point between them is at a distance of $ 20cm $ from the $ 20coulomb $ charge. Charge $ Q $ is

Options:

A) $ 30C $

B) $ 40C $

C) $ 60C $

D) $ 80C $

Show Answer

Answer:

Correct Answer: D

Solution:

At neutral point $ k\times \frac{20}{{{(20\times {{10}^{-2}})}^{2}}}=k\times \frac{Q}{{{(40\times {{10}^{-2}})}^{2}}} $

therefore Q = 80 C



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