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At what rate must the potential difference between the plates of a parallel-plate capacitor with a2.0μF capacitance be changed to produce a displacement current of1.5A?

Short Answer

Expert verified

The rate of potential is 7.5×105V/s.

Step by step solution

01

Given data

C=2.0μF

id=1.5A

02

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,Ï• is the flux.

03

Determining the rate of change of potential difference

From the equation 32-10,

id=ε0dϕEdt

Where, Ï•=AEtheand theE=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=7.5×105V/s

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

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