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The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with ε = 80 ε0. The surface charge density on the conductor is—
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- 0 c/m2
- 2 c/m2
- 1.8 × 10– 11 c/m2
- 1.41 × 10– 9 c/m2
Correct Option: D
Electric field is given by
| E = | |
| ε0 |
for the system to be immersed in water its dielectric changes form ε0 to 80 ε0 so we get surface charge density as
| σ = ε0εr|→E| ⇒ 80ε0 E = | × 2 | |
| 36π |
= 1.41 × 10–9.
