Two particles A and B having charges of $4 \times 10^{- 6} C and - 8 \times 10^{- 6} C$ respectively are held fixed at a separation of 60 cm. Locate the point(s) on the line AB where the electric potential is zero.

1.6

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Solution

The correct option is B 0.6

Two particles $A$ and $B$ having charges $= 4 \times 10^{- 6} C$ and $- 8 \times 10^{- 6} C$ .
$d = 60 c m = \frac{60}{100} = 0.6 m$
$\frac{k (4 \times 10^{- 6} \times Q)}{(x + 0.6)} = \frac{k (8 \times 10^{- 6}) \times Q}{x}$
or, $x^{2} = 2 {(x + 0.6)}^{2}$
or, $x^{2} = 2 {(x^{2} + 2.06 \times x + 0.6)}^{2}$
or, $x^{2} = {2 x}^{2} + 4.12 x + 0.72$
or, $x^{2} + 4.12 x + 0.72 = 0$
or, $\frac{1}{x + 0.6} = \frac{2}{x}$
or, $x = 2 x + 0.12$
or, $2 x = - 0.12$
or, $x = - 0.6 m$

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Two particles A and B having charges of 4×10−6Cand−8×10−6C respectively are held fixed at a separation of 60 cm. Locate the point(s) on the line AB where the electric potential is zero.

The correct option is B 0.6Two particles A and B having charges =4×10−6C and −8×10−6C.d=60cm=60100=0.6mk(4×10−6×Q)(x+0.6)=k(8×10−6)×Qxor, x2=2(x+0.6)2or, x2=2(x2+2.06×x+0.6)2or, x2=2x2+4.12x+0.72or, x2+4.12x+0.72=0or, 1x+0.6=2xor, x=2x+0.12or, 2x=−0.12or, x=−0.6m

Two particles A and B having charges of $4 \times 10^{- 6} C and - 8 \times 10^{- 6} C$ respectively are held fixed at a separation of 60 cm. Locate the point(s) on the line AB where the electric potential is zero.