Name: Electric Field of a Sphere
Uploaded: 2019-07-19
Duration: 5 min 10 s
Description: This video demonstrates how an electric field varies as you move away from a point charge or a uniform charge distribution on a sphere. Watch this video.

Question

Apply Gauss's law to derive the expression for electric field intensity due to an infinitely long straight uniformly charged wire. What is the direction of the field intensity it is positively charged?

Accepted Answer

According to Gauss's law, the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Gauss's law is true for any closed surface, no matter what its shape or size.

$\int \vec{_{s} E} d \vec{S} cosθ = \frac{q}{ε_{0}}$

Consider an infinitely long thin straight wire with uniform linear charge $q$ and density $λ$ . Let $P$ be a point at distance $r$ from the wire. Draw a cylindrical surface of radius $r$ with point $P$ on its surface. The cylindrical surface is a Gaussian surface.

The total charge enclosed by Gaussian surface due to continuous charge distribution along the length $l$ of the wire is given by, $q = λl$ . From Gauss's law, we have,

$\int_{s} E d \vec{S} cosθ = \frac{q}{ε_{0}} = \frac{λl}{ε_{0}} or E \int_{s} dS = \frac{λl}{ε_{0}}$ as for the curved surface, the angle between the direction of the field and the normal to the surface is zero. As the total surface area $\int dS = 2 πrl$ , we get,

$E (2 πrl) = \frac{λl}{ε_{0}} or E = \frac{1}{2 {πε}_{0}} . \frac{λ}{r}$ .

The direction of the field intensity will be vertically outwards of the surface of the wire.

Apply Gauss's law to derive the expression for electric field intensity due to an infinitely long straight uniformly charged wire. What is the direction of the field intensity it is positively charged?

Important Questions on Electric Charges and Fields

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