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Define electric flux. Write its SI …

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Define electric flux. Write its SI unit. Gauss' law in electrostatics is true for any closed surface, no matter what its shape or size is. Justify this statement with the help of a suitable example. (ii) Use Gauss' law to prove that the electric field inside a uniformly charged spherical shell is zero.

Posted by Maahnoor Quraishi 2 years, 4 months ago

CBSE > Class 12 > Physics

2 answers

Preeti Dabral 2 years, 4 months ago

The electric flux through an area is defined as the electric field multiplied by the area of the surface projected on a plane, perpendicular to the field. Its S.I. unit is voltmeter (V_m) or Newton metre square per coulomb (Nm² C^-1). The given statement is justified because while measuring the flux, the surface area is more important than its volume on its size.
Electric field inside the shell:

The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let's consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude E_i on the Gaussian surface and is directed radially outward.
Electric flux through the Gaussian surface is given by,
$=\int_{s} \vec{E}_{i} \cdot d \vec{S}$
$=\int E_{i} d S \cos 0=E_{i} .4 \pi r^{2}$
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.
Therefore, using Gauss's theorem, we have
$\int_{S} \vec{E}_{i} \cdot d \vec{S}=\frac{1}{\epsilon_{0}} \times \text { charge enclosed }$
$\Rightarrow E_{i} \cdot 4 \pi r^{2}=\frac{1}{\epsilon_{0}} \times 0$
$\Rightarrow$ E_i = 0
Thus, electric field at each point inside a charged thin spherical shell is zero.

Mayank Chauhan 2 years, 4 months ago

The no. Of electric field lines passing through a area vector perpendicularly Its S.I unit is Nm²per C

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