Two point charges of 4 microcoulombs and -2 microcoulombs are separated by a distance of 1 meter in air. At what distance from the 4 microcoulomb charge on the line joining them will the electric potential be zero?

C. 0.67 m — The electric potential is zero when the sum of potentials from both charges is zero. Setting k(q_1/x) + k(q_2/(1-x)) = 0 and solving for x gives x = 2/3 meters.

JEE Advanced · PhysicsMedium

$Two point charges of q_{1} = 4 \times 1 0^{- 6} C and q_{2} = - 2 \times 1 0^{- 6} C are separated by a distance of 1 m . At what distance x from q_{1} on the line joining them will the electric potential be zero?$

A. $0.33 m$
B. $0.50 m$
C. $0.67 m$
D. $0.75 m$

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Correct answer: C — $0.67 m$

Solution

The potential V = k (q_{1} / x + q_{2} / (d - x)) = 0 . Substituting values, $4/x = 2/(1-x)$, which simplifies to $4 - 4x = 2x$, yielding x = 2/3 \approx 0.67 m .

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Two point charges of q1​=4×10−6 C and q2​=−2×10−6 C are separated by a distance of 1 m. At what distance x from q1​ on the line joining them will the electric potential be zero?

Solution

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More Electrostatics practice questions

$Two point charges of q_{1} = 4 \times 1 0^{- 6} C and q_{2} = - 2 \times 1 0^{- 6} C are separated by a distance of 1 m . At what distance x from q_{1} on the line joining them will the electric potential be zero?$