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Determine the rate at which the electric field changes between the round plates of a capacitor, 8.0 cm in diameter, if the plates are spaced 1.4 mm apart and the voltage across them is changing at a rate of 110 V/s .

User Sanusart
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1 Answer

4 votes

Solution:

The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :


$E=(V)/(D)$

Differentiating on both the sides with respect to time, we get


$(dE)/(dt)=(1)/(D)(dV)/(dt)$

Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :


$(dE)/(dt)=(1)/(D)(dV)/(dt)$


$=(1)/(1.4 * 10^(-3)) * 110$


$=7.85 * 10^4$ V/m-s

User Remek Ambroziak
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