Skip to content

Question

An electric dipole with moment \( p = 2 \times 10^{-9} \, \text{C m} \) lies along the z-axis. What is
the potential at \( (0, 0, 4) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,
\text{Nm}^2 \text{C}^{-2} \)).

Options

Choose one · Correct answer highlighted

Explanation

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (2 × 10⁻⁹/4²) = 9 × 10⁹ × (2 × 10⁻⁹/16) = 1.125 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.