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Standard XII

Physics

Gauss's Law

The electric ...

Question

The electric field on two sides of a charged plate is shown in the figure. The surface charge density on the plate is given by :

2 E_{0}

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{4 E}_{0}

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{10 E}_{0}

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Zero

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Solution

The correct option is B ${4 E}_{0}$
Solution Net Electric flux
$=_{1} ._{1} +_{2} ._{2}$
$E_{1} = 8 v / m$
$= E_{1} . A c o s (180^{\circ}) + E_{2} A c o s (0^{\circ})$
$= (8) A (- 1) + (12) A (1)$
$= 4 A$
By Gauss Law :
Net flux = $\frac{g i n c l o s e d}{ε_{0}}$
$\Rightarrow 4 A = \frac{g i n c l o s e d}{ε^{0}}$
$\Rightarrow 4 ε_{0} = \frac{g i n c l o s e d}{A} =$ surface charge density of plot.
Option - B is correct.

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The electric field on two sides of a charged plate is shown in the figure. The surface charge density on the plate is given by :

The correct option is B 4 E0Solution Net Electric flux=→E1.→n1+→E2.→n2 E1=8v/m=E1.Acos(180∘)+E2Acos(0∘)=(8)A(−1)+(12)A(1)=4ABy Gauss Law :Net flux = ginclosedε0⇒4A=ginclosedε0⇒4ε0=ginclosedA= surface charge density of plot.Option - B is correct.