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Byju's Answer

Standard XII

Physics

Parallel and Series Combination of Capacitors

A parallel pl...

Question

A parallel plate capacitor with plates of area $1 m^{2}$ each, area $t$ a separation of $0.1 m$ . If the electric field between the plates is $100 N / C$ , the magnitude of charge each plate is:-
$(T a k e ε_{0} = 8.85 \times 10^{- 12} \frac{c^{2}}{N - m^{2}})$

7.85 \times 10^{- 10} C

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6.85 \times 10^{- 10} C

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9.85 \times 10^{- 10} C

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8.85 \times 10^{- 10} C

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Solution

The correct option is D $8.85 \times 10^{- 10} C$
$E = \frac{σ}{\in_{0}} = \frac{Q}{A \in_{0}}$
$Q = A E \in_{0}$
$Q = (1) (100) (8.85 \times 10^{- 12})$
$Q = 8.85 \times 10^{- 19} C$

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Similar questions

Q. A parallel plate capacitor with plates of area

1 m^{2}

each, are at a separation

0.1 m

. If the electric field between the plates is

100 {NC}^{- 1}

, the magnitude of charge on each plate is:

(Take ε_{0} = 8.85 \times 10^{- 12} \frac{C^{2}}{{Nm}^{2}})

Each plate of a parallel plate air capacitor has area

S = 5 \times 10^{- 3} m^{2}

and the distance between the plates is

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. Plate

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q_{1} = + 10^{- 10} C

, and plate

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q_{2} = + 2 \times 10^{- 10} C

. A battery of emf

E = 10 V

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and the negative terminal to plate

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. Charge supplied by the battery is :

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ε_{0} = 8.85 \times 10^{- 12} N m^{2} C^{- 2}

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Q. A parallel plate capacitor with area

200 {c m}^{2}

and separation between the plates

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(ϵ_{0} = 8.85 \times 10^{- - 12} \frac{C^{2}}{N . m^{2}})

Q. Magnitude of surface charge density on each plate of a parallel plate capacitor is

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. What is the magnitude of the electric field in the region between the plates?
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ϵ_{o} = 8.85 \times 10^{- 12} C^{2} {/N m}^{2}

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The breakdown field strength for dry air is $3 \times 10^{6}$ V/m. Each of the plates of the parallel plate capacitor has an area of $0.2 m^{2}$ . The maximum charge that can be collected is : (Given permittivity of air $ϵ_{\circ} = 8.85 \times 10^{- 12} C^{2} / N m^{2}$ )

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A parallel plate capacitor with plates of area 1m2 each, area t a separation of 0.1m. If the electric field between the plates is 100N/C, the magnitude of charge each plate is:-(Takeε0=8.85×10−12c2N−m2)

The correct option is D 8.85×10−10CE=σ∈0=QA∈0Q=AE∈0Q=(1)(100)(8.85×10−12)Q=8.85×10−19C

A parallel plate capacitor with plates of area $1 m^{2}$ each, area $t$ a separation of $0.1 m$ . If the electric field between the plates is $100 N / C$ , the magnitude of charge each plate is:-
$(T a k e ε_{0} = 8.85 \times 10^{- 12} \frac{c^{2}}{N - m^{2}})$

The correct option is D $8.85 \times 10^{- 10} C$
$E = \frac{σ}{\in_{0}} = \frac{Q}{A \in_{0}}$
$Q = A E \in_{0}$
$Q = (1) (100) (8.85 \times 10^{- 12})$
$Q = 8.85 \times 10^{- 19} C$