Electric field intensity at a point due to a charge Q is 24 N/C and electric potential at A' due to charge Q is =12 j/C The distance between the charges and magnitude of Q are

0.4 m, 0.667 \times 10^{- 9} C

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0.4 m, 6.67 \times 10^{- 4} C

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0.5 m, 2.67 \times 10^{- 9} C

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0.5 m, 7.66 \times 10^{- 9} C

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Solution

The correct option is C $0.5 m, 2.67 \times 10^{- 9} C$
$E = \frac{K Q}{A B^{2}} = 24 N / C$
$V = \frac{K Q}{A B} = 12 J / C$
So $,$ $A B = \frac{V}{E} = \frac{12}{24} = 0.50 m$
So $,$ $V = \frac{9 \times 10^{9} \times Q}{0.50} = 12 J / C$
$Q = 2.67 \times 10^{- 9} C$
Hence,
option $(C)$ is correct answer.

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Electric field intensity at a point due to a charge Q is 24 N/C and electric potential at A' due to charge Q is =12 j/C The distance between the charges and magnitude of Q are

The correct option is C 0.5m,2.67×10−9CE=KQAB2=24N/CV=KQAB=12J/CSo, AB=VE=1224=0.50mSo, V=9×109×Q0.50=12J/CQ=2.67×10−9CHence,option (C) is correct answer.