Question

Some equipotential surfaces are shown. The electric field at any point is

• A. $\frac{6}{x^{2}}$perpendicular to the plane of paper
• B. $\frac{600}{x^{2}}$perpendicular to the plane of paper
• C. $\frac{6}{x^{2}}$radially
• D. $\frac{600}{x^{2}}$radially

Answer 1

Answer: (C)

$\frac{6}{x^{2}}$radially

Answer 2

V = $\frac { \mathrm { q } } { 4 \pi \in _ { 0 } r }$ Ž q = 4pĪ₀Vr = 60 × 0.1 × 4pĪ₀ = 24pĪ₀

\ E = $\frac { \mathrm { q } } { 4 \pi \epsilon _ { 0 } \mathrm { r } ^ { 2 } } = \frac { 24 \pi \epsilon _ { 0 } } { 4 \pi \epsilon _ { 0 } \mathrm { r } } = \frac { 6 } { \mathrm { x } ^ { 2 } }$

Direction of electric field is radially outward.