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Q.

Two particles A and B (B is right of A) having charges $8 \times 10^{- 6} C a n d - 2 \times 10^{- 6} C,$ respectively, are held fixed with separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?

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a

5 cm right of B

b

5 cm left of A

c

20 cm left of A

d

20 cm right of B

answer is D.

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Detailed Solution

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Let the third charge Q be placed at a distance x to the right of B. Then

$\frac{kQ \times 8 \times 10^{- 6}}{{(20 + x)}^{2}} = \frac{kQ (2 \times 10^{6})}{x^{2}}$

So, x = 20 cm

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Two particles A and B (B is right of A) having charges 8×10−6 C and −2×10−6 C, respectively, are held fixed with separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force?