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Fundamentals Of Physics
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Short Answer

At what rate must the potential difference between the plates of a parallel-plate capacitor with a2.0 μF capacitance be changed to produce a displacement current of 1.5 A?

The rate of potential is 7.5×105 V/s.

See the step by step solution

Step by Step Solution

Given data

C=2.0 μF

id=1.5 A

Determining the concept

By using the concept of displacement current and by using equation 32-10, write the equation for flux as a function of potential, and from that, find the rate of change of potential difference. The rate of change of potential difference across the plates of the capacitor is displacement current per unit capacitance.

Formulae are as follows:

id=ε0dϕEdt

where, 'i' is current, ϕ is the flux.

Determining the rate of change of potential difference

From the equation 32-10,

id=ε0dϕEdt

Where, ϕ=AEthe and the E=Vd.

So,

ϕE=AVd

So,

id=Aε0ddVdt

Where, the capacitance,

C=Aε0d

So, the rate of potential is,

dVdt=idCdVdt=1.52.0×10-6dVdt=7.5×105 V/s

Hence, the rate of potential is 7.5×105 V/s.

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