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Question: A parallel plate capacitor with plates of area \(1{\rm{ }}{{\rm{m}}^2}\) each, area \(t\) a separati...

A parallel plate capacitor with plates of area 1m21{\rm{ }}{{\rm{m}}^2} each, area tt a separation of 0.1m0.1{\rm{ m}}. If the electric field between the plates is 100N/C100{\rm{ N/C}}, the magnitude of charge each plate is:-
(Take ∈0=8.85×10−12c2Nm2{ \in _0} = 8.85 \times {10^{ - 12}}\dfrac{{{c^2}}}{{N{m^2}}})
A. 7.85×10−10C7.85 \times {10^{ - 10}}{\rm{ C}}
B. 6.85×10−10C6.85 \times {10^{ - 10}}{\rm{ C}}
C. 9.85×10−10C9.85 \times {10^{ - 10}}{\rm{ C}}
D. 8.85×10−10C8.85 \times {10^{ - 10}}{\rm{ C}}

Explanation

Solution

The electric field inside a parallel plate capacitor is given by, E=σ∈0=Q∈0AE = \dfrac{\sigma }{{{ \in _0}}} = \dfrac{Q}{{{ \in _0}A}} where, σ\sigma is the surface charge density, Q is the charge, A is the area of the plates and ∈0{ \in _0} is the permittivity of free space.

Complete step by step solution:
Given,
The area of capacitor plate, A=1m2A = 1{\rm{ }}{{\rm{m}}^2}
The separation of in between two plates, t=0.1mt = 0.1{\rm{ m}}
The electric field between the plates, E=100N/CE = 100{\rm{ N/C}}
The permittivity of free space, ∈0=8.85×10−12c2Nm2{ \in _0} = 8.85 \times {10^{ - 12}}\dfrac{{{c^2}}}{{N{m^2}}}.
When we find the electric field between the plates of a parallel plate capacitor we assume that the electric field from both plates is,
E=σ2∈0E = \dfrac{\sigma }{{2{ \in _0}}} …… (1)
The factor two in the denominator of equation (1) is due to the fact that there is a surface charge density on both the sides of the plate. Due to electric field, the charges accumulate on the surface of the plates depending on the polarity. Therefore, when both the plates are put together, the net electric field is given by,

E=σ∈0E = \dfrac{\sigma }{{{ \in _0}}} …… (2)
Now the surface charge density is the charge per area in a capacitor plate and is given by,
σ=QA\sigma = \dfrac{Q}{A} …… (3)
Substituting value of σ\sigma in equation (2) we get,
E=σ∈0=Q∈0AE = \dfrac{\sigma }{{{ \in _0}}} = \dfrac{Q}{{{ \in _0}A}} …… (4)

Substituting the value of 1m21{\rm{ }}{{\rm{m}}^2} in AA, 100N/C100{\rm{ N/C}} in EE, 8.85×10−12C2Nm28.85 \times {10^{ - 12}}\dfrac{{{C^2}}}{{N{m^2}}} in ∈0{ \in _0} in equation (4) we get,

E=Q∈0A ⇒Q=∈0AE=8.85×10−12×1×100=8.85×10−10C\begin{array}{l}E = \dfrac{Q}{{{ \in _0}A}}\\\ \Rightarrow Q = { \in _0}AE = 8.85 \times {10^{ - 12}} \times 1 \times 100 = 8.85 \times {10^{ - 10}}{\rm{ C}}\end{array}

Hence, the correct option is (D).

Note: The students have to apply the concept of charge in parallel plate capacitor and the electric field developed when voltage is applied. The calculation of charge in parallel plate capacitor can be done using the relation of an electric field with the area of the plate and the permittivity of free space.