Question

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.
Figure

Answer 1

Answer:

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Answer 2

a) The charge on the outer surface of the upper plate is 0.5 µC .

b) The potential difference developed between the upper and the middle plates is 10V .

Explanation:

a) When charge of 1μc is induced in to the plate B, then the charge on the upper surface of the plate A is 0.5 µC

b) From the information given in the question is C = 50 µF

$=50 \times 10-9=5 \times 10-8=0.5 \times 10-6 c$

We know that, potential difference (V) = Q/C

$\Rightarrow 5 \times 10-7 / 5 \times 10-8=10 v$

Hence, we can conclude that the potential difference between the two plates of the capacitor is 10V.