Charges +Q and −Q, with Q = 2.00 nC, lie at x = +0.0500 m and x = −0.0500 m respectively. At x = +0.200 m calculate electric potential and field. Find the work done by the electric field when a +1.00 nC test charge moves slowly from x = 0.200 m to x = 0.300 m. Derive the far-axis approximations for V and E in terms of dipole moment p = Qd and compare the exact potential at 0.200 m with the approximation.

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At 0.200 m, V = 47.9 V and E = 512 N C⁻¹ toward +x. At 0.300 m, V = 20.5 V, so field work = +2.74 × 10⁻⁸ J. Far away V ≈ kp/x² and E ≈ 2kp/x³; the potential approximation at 0.200 m is 44.9 V. At x = 0.200 m, distances are 0.150 m to +Q and 0.250 m to −Q. Potential is kQ(1/0.150−1/0.250) = 47.9 V. Both axial field contributions point toward +x there: E = kQ[1/(0.150)²−1/(0.250)²] = 512 N C⁻¹. At 0.300 m, V = kQ(1/0.250−1/0.350) = 20.5 V. Electric-field work is q(V_i−V_f) = 1.00 × 10⁻⁹(47.9−20.5) = 2.74 × 10⁻⁸ J. For x much larger than separation d, expand 1/(x∓d/2) to first order. Their difference is approximately d/x², so V ≈ kp/x² and E = −dV/dx ≈ 2kp/x³. Here p = Qd = 2.00 × 10⁻¹⁰ C m, giving V_approx = 44.9 V. Its roughly 6% underestimate shows that x/d = 2 is only moderately within the far-field regime.

Electric potential along a dipole axis

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