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Standard XII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

A parallel pl...

Question

A parallel plate capacitor is charged to $60 μ C$ . Due to a radioactive source, the plate loses charge at the rate of $1.8 \times 10^{- 8} C / s$ . The magnitude of displacement current is:

1.8 \times 10^{- 8} C / s

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3.6 \times 10^{- 8} C / s

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4.1 \times 10^{- 11} C / s

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5.7 \times 10^{- 12} C / s

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Solution

The correct option is A $1.8 \times 10^{- 8} C / s$
The displacement current is that current which comes into play in the region in which the electric field and hence the electric flux is changing with time.
Maxwell found that conduction current (I) and displacement current $(I_{d})$ together have the property of continuity, although individually they may not be continuous. Maxwell also predicted that this current produces the same magnetic field as a conduction current can produce.
Displacement current is given by
$I_{d} = \frac{d q}{d t} = 1.8 \times 10^{- 8} C / s$

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A parallel plate capacitor is charged to 60μC. Due to a radioactive source, the plate loses charge at the rate of 1.8×10−8C/s. The magnitude of displacement current is:

A parallel plate capacitor is charged to $60 μ C$ . Due to a radioactive source, the plate loses charge at the rate of $1.8 \times 10^{- 8} C / s$ . The magnitude of displacement current is: