Each plate of a parallel plate air capacitor has area A=5×10−3m2 and are d=8.85 mm apart as shown in figure. Plate A has positive charge q1=10−10 coulomb and plate B has charge q2=2×10−10 coulomb. When its positive terminal of battery of emf 10 volts is connected with plate. A and negative terminal with plate B.Choose the correct statement(s)

10=(q−0.5×10−10)dAε0 q=10−10C

Each plate of a parallel plate air capacitor has area A=5×10−3m2 and are d=8.85 mm apart as shown in figure. Plate A has positive charge q1=10−10 coulomb and plate B has charge q2=2×10−10 coulomb. When its positive terminal of battery of emf 10 volts is connected with plate. A and negative terminal with plate B.Choose the correct statement(s)

10=(q−0.5×10−10)dAε0 q=10−10C

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Each plate of a parallel plate air capacitor has area $A = 5 \times 10^{- 3} m^{2}$ and are $d = 8.85 m m$ apart as shown in figure. Plate A has positive charge $q_{1} = 10^{- 10}$ coulomb and plate B has charge $q_{2} = 2 \times 10^{- 10}$ coulomb. When its positive terminal of battery of emf 10 volts is connected with plate. A and negative terminal with plate B.
Choose the correct statement(s)

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The amount of charge on left surface of plate A will decrease when the battery is connected.

The amount of charge passes through the battery is $2 \times 10^{- 10} C$ .

The amount of charge on right surface of plate B will not change when the battery is connected.

The amount of charge passes through the battery is $1 \times 10^{- 10} C$ .

answer is B, D.

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Detailed Solution

$10 = \frac{(q - 0.5 \times 10^{- 10}) d}{A ε_{0}} q = 10^{- 10} C$

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