The electrical potential 2.0 cm from a point charge of Q₁ = +6.5 × 10^-9 C and 3.5 cm from a second charge Q₂ is 1.2 kV. Find Q₂.

Question

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The potential due to a point charge is given by V = kQ/r, where k is Coulomb's constant, Q is the charge, and r is the distance from the charge. In this problem, we have two charges, Q1 and Q2, and we are given the potential at two points, r1 = 2.0 cm and r2 = 3.5 cm. We can use these equations to solve for Q2:V = kQ1/r1 + kQ2/r21.2 kV = (9 × 10^9 Nm^2/C^2)(6.5 × 10^-9 C)/0.02 m + (9 × 10^9 Nm^2/C^2)(Q2)/0.035 m1.2 kV = 2.925 × 10^3 V + 2.571 × 10^3 V/Q2Q2 = -6.7 × 10^-9 C

The Electrical Potential 2