Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Show that the electric field at the surface of a charged conducting sphere is given by

\vec{E} = \frac{σ}{ε_{0}} \hat{n}

, where

σ

is the surface charge density and

\hat{n}

is a unit vector normal to the surface in the outward direction.

Accepted Answer

Let a charge $Q$ be given to a conductor, this charge under electrostatic equilibrium will redistribute and the electric field inside the conductor is zero (i.e., $E_{i n} = 0$ ).

Let us consider a point P at which electric field strength is to be calculated, just outside the surface of the conductor. Let the surface charge density on the surface of the conductor in the neighbourhood of P be $\hat{n}$ . Now consider a small cylindrical box CD having one base C passing through P; the other base D lying inside the conductor and the curved surface being perpendicular to the surface of the conductor.

Let the area of each flat base be $a$ . As the surface of the conductor is equipotential surface, the electric field strength E at P, just outside the surface of the conductor is perpendicular to the surface of the conductor in the neighbourhood of P. The flux of electric field through the curved surface of the box is zero, since there is no component of electric field E normal to curved surface. Also, the flux of electric field through the base D is zero, as electric field strength inside the conductor is zero. Therefore, the resultant flux of electric field through the entire surface of the box is same as the flux through the face C. This may be analytically seen as:

If $S_{1}$ and $S_{2}$ are flat surfaces at C and D and $S_{3}$ is curved surface, then,

Total electric flux,

$\int_{S} \vec{E} . \vec{d S} = \int_{S_{1}} \vec{E} . \vec{d S_{1}} + \int_{S_{2}} \vec{E} . {\vec{d S}}_{2} + \int_{S_{3}} \vec{E} . {\vec{d S}}_{3} = \int_{S_{1}} E . d S_{1} \cos 0^{\circ} + \int_{S_{2}} 0 . {\vec{d S}}_{2} + \int_{S_{3}} E . d S_{3} \cos 90^{\circ} = \int_{S_{1}} E . d S_{1} = E a$

As the charge enclosed by the cylinder is ( $σ a$ ) coulomb, we have, using Gauss’s theorem,

Total electric flux = $\frac{1}{\in_{\circ}} \times$ Charge enclosed

Thus, the electric field strength at any point close to the surface of a charged conductor of any shape is equal to $\frac{1}{\in_{\circ}}$ times the surface charge density $σ$ . This is known as Coulomb’s law. The electric field strength is directed radially away from the conductor if $σ$ is positive and towards the conductor if $S$ is negative.

If $\hat{n}$ is unit vector normal to surface in outward direction, then

$\vec{E} = \frac{σ}{\in_{\circ}} \hat{n}$

Obviously electric field strength near a plane conductor is twice of the electric field strength near a non-conducting thin sheet of charge.

Show that the electric field at the surface of a charged conducting sphere is given by $\vec{E} = \frac{σ}{ε_{0}} \hat{n}$ , where $σ$ is the surface charge density and $\hat{n}$ is a unit vector normal to the surface in the outward direction.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

Show that the electric field at the surface of a charged conducting sphere is given by E→=σε0n^, where σ is the surface charge density and n^ is a unit vector normal to the surface in the outward direction.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

Show that the electric field at the surface of a charged conducting sphere is given by $\vec{E} = \frac{σ}{ε_{0}} \hat{n}$ , where $σ$ is the surface charge density and $\hat{n}$ is a unit vector normal to the surface in the outward direction.