Two point charges, q1 = +5.0 μC and q2 = -2.0 μC, are placed 0.30 m apart in vacuum. What is the electric potential at the midpoint between the charges? (Use k = 9.0 × 10^9 N·m^2/C^2)
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Correct answer
A. 1.8 × 10^5 V
Principle or equation
The electric potential due to a point charge is V = kQ/r, where r is the distance from the charge. For multiple charges, the total potential is the algebraic sum of the individual potentials (since potential is a scalar).
Why this answer is correct
At the midpoint, each charge is 0.15 m from the point. The potential due to q1 is V1 = (9.0 × 10^9)(5.0 × 10^-6)/0.15 = 3.0 × 10^5 V. The potential due to q2 is V2 = (9.0 × 10^9)(-2.0 × 10^-6)/0.15 = -1.2 × 10^5 V. The total potential is V = V1 + V2 = 3.0 × 10^5 - 1.2 × 10^5 = 1.8 × 10^5 V.
Example
If q1 = +4 μC and q2 = -1 μC are 0.20 m apart, the potential at the midpoint (r = 0.10 m) is (9e9)(4e-6)/0.10 + (9e9)(-1e-6)/0.10 = 3.6e5 - 0.9e5 = 2.7e5 V.
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