The electric field near a conducting surface | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Gauss's Law,for a charged conductor,the electric field just outside the surface is perpendicular to the surface.By applying Gauss's L

Vedclass · Accepted Answer

$\frac{\sigma}{\varepsilon_0}$ and is normal to the surface

Vedclass · Answer

(C) According to Gauss's Law,for a charged conductor,the electric field just outside the surface is perpendicular to the surface.By applying Gauss's Law to a small cylindrical Gaussian surface enclosing a charge $\sigma A$ on the conductor,we have $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.Since the electric field inside the conductor is zero and the field is perpendicular to the surface,the flux through the curved surface is zero,and the flux through the inner flat face is zero.Thus,$E \cdot A = \frac{\sigma A}{\varepsilon_0}$,which simplifies to $E = \frac{\sigma}{\varepsilon_0}$.Therefore,the electric field is $\frac{\sigma}{\varepsilon_0}$ and is directed normal to the surface.

The electric field near a conducting surface having a uniform surface charge density $\sigma$ is given by:

Similar Questions

The electric potential in a region around the origin is given by $V(x) = 4x^2$. The total charge enclosed in a cube of side $1 \ m$ centered at the origin is (in Coulombs) ...

$A$ spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

Find the ratio of the electric field at points $A$ and $B$. An infinitely long uniformly charged wire with linear charge density $\lambda$ is kept along the $z$-axis.

Mention applications of Gauss's law.

The electrostatic field of a long uniformly charged wire varies with distance $r$ according to which relation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module