## Related questions with answers

A certain parallel-plate capacitor is filled with a dielectric for which κ = 5.5. The area of each plate is 0.034 m², and the plates are separated by 2.0 mm.The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 200 kN/C. What is the maximum energy that can be stored in the capacitor?

Solutions

VerifiedThis is a complex problem in which we need to find the maximal energy that can be stored in the given capacitor. We need to use the relation of the stored energy, voltage and capacitance in this problem. We know that the equation is:

$U=\frac{1}{2}\cdot C\cdot V^2\tag1$

But how do we find the capacitance and voltage? Keep in mind that voltage is the same thing as potential difference.

capacitance:

$C = k\ \epsilon_0 \ A/d = (5.5)*(8.85e-12)*(0.034)/(2.0e-3) = 827.475 \ pF$

maximum potential difference:

$V = E \ d = (200e3) * (2e-3) = 400 \ V$

maximum stored energy:

$E = (1/2)\ C\ V^2 = (1/2) * (827.475e-12) *(400)^2 = 66 \ \mu J$

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