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Question

Charge between parallel plates :

Surface charge density is defined as the charge per unit surface area of surface charge distribution. i.e., σ=qS

Two large thin metal plates are parallel and close to each other, on their inner faces, the plates have surface charge densities of opposite sign having magnitude of 1.70×1022 Cm2 as shown in figure.
[Use ε0=8.85×1012 Fm1]

(i) The electric field E in the region in between the plates A and B is



A
12×1012 NC1
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B
2×1010 NC1
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C
0.2×1010 NC1
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D
1.2×1012 NC1
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Solution

The correct option is C 0.2×1010 NC1



Given, surface charge densities of plate <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> A and B is,

σA=+1.70×1022 Cm2 ; σB=1.70×1022 Cm2

Using gauss's law, electric field in the region between the plate A and B is,

E=EA+EB=σ2ε0+σ2ε0=σε0

E=σε0=1.70×10228.85×1012

E0.2×1010 NC1

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (C) is the correct answer.


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