Electric field intensity at a point B due to a point charge Q kept at a point charge Q kept at point A is `24 NC^(-1)`, and electric potential at B due to the same charge is `12 JC^(-1)`. Calculate the distance AB and magnitude of charge.

Electric field intensity at a point B due to a point charge Q kept at

1.	Electric field intensity at a point B due to a point charge Q kept at a point charge Q kept at point A is `24 NC^(-1)`, and electric potential at B due to the same charge is `12 JC^(-1)`. Calculate the distance AB and magnitude of charge.
Answer» Here, `E = (Q)/(4pi in_(0) r^(2)) = 24 NC^(-1)`, and `V = (Q)/(4pi in_(0) r) = 12 JC^(-1)` Dividing, we get `(V)/(E) = r = (12)/(24) = 05m = AB` From `V = (Q)/(4pi in_(0) r) , Q = 4pi in_(0) r xx V` `Q = (1)/(9xx10^(9)) xx 05xx12 = 0*667 xx 10^(-9) C`

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Electric field intensity at a point B due to a point charge Q kept at a point charge Q kept at point A is `24 NC^(-1)`, and electric potential at B due to the same charge is `12 JC^(-1)`. Calculate the distance AB and magnitude of charge.

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