The capacitance between the adjacent plates shown in figure (31-E20) is 50 nF. A charge of $1.0 μ C$ is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate ? (b) Find the potential difference developed between the upper and the middle plates.

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Solution

(a) When charge of $1 μ C$ is introduced to the B plate, we also get a $0.5 μ C$ charge on the upper surface of plate A

(b) Given, $C = 50 μ F = 50 \times 10^{- 9} F$

$= 5 \times 10^{- 8} F$

Now charge $= 0.5 \times 10^{- 6} C$

$v = 10 V$

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The capacitance between the adjacent plates shown in figure (31-E20) is 50 nF. A charge of 1.0 μC is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate ? (b) Find the potential difference developed between the upper and the middle plates.

(a) When charge of 1 μC is introduced to the B plate, we also get a 0.5 μC charge on the upper surface of plate A (b) Given, C=50 μF=50×10−9 F =5×10−8 F Now charge =0.5×10−6 C v=10 V

The capacitance between the adjacent plates shown in figure (31-E20) is 50 nF. A charge of $1.0 μ C$ is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate ? (b) Find the potential difference developed between the upper and the middle plates.

(a) When charge of $1 μ C$ is introduced to the B plate, we also get a $0.5 μ C$ charge on the upper surface of plate A

(b) Given, $C = 50 μ F = 50 \times 10^{- 9} F$

$= 5 \times 10^{- 8} F$

Now charge $= 0.5 \times 10^{- 6} C$

$v = 10 V$