The electric field at the surface of a conductor is parallel...

The electric field at the surface of a conductor is parallel to the surface. depends only on the total charge on the conductor. depends only on the area of the conductor. depends only on the curvature of the surface. depends on the area and curvature of the conductor and on its charge.

Let us mentally impart charges $+q$ and $-q$ to the conductors. As the medium is poorly conducting, the surfaces of the conductors are equipotential and the field configuration is same as in the absence of the medium. Let us surround, for example, the positively charged conductor, by a closed surface $S$, just containing the conductor, then, $\quad R=\frac{\varphi}{i}=\frac{\varphi}{\int \vec{j} \cdot d \vec{S}}=\frac{\varphi}{\int \sigma E_{n} d S} ;$ as $\overrightarrow{j \uparrow \uparrow} \vec{E}$ and $\quad C=\frac{q}{\varphi}=\frac{\varepsilon \varepsilon_{0} \int E_{n} d S}{\varphi}$ So, $R C=\frac{\varepsilon \varepsilon_{0}}{\sigma}=\rho \varepsilon \varepsilon_{0}$

Solutions to I.E. Irodov's problems in general physics

Electrodynamics

Electric Current

Question

The electric field at the surface of a conductor is parallel to the surface. depends only on the total charge on the conductor. depends only on the area of the conductor. depends only on the curvature of the surface. depends on the area and curvature of the conductor and on its charge.

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