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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Application +4 | -1
Force of 4000 N4000 \mathrm{~N} acts on 4 C4 \mathrm{~C} charge placed in a uniform electric field. The potential difference between two points 2 cm2 \mathrm{~cm} apart in the field is
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Correct option: D
20 V20 \mathrm{~V}
Option A (6×10−6 T6 \times 10^{-6} \mathrm{~T}) is correct because the magnetic field is calculated using the formula for the magnetic field around a straight wire.

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