A parallel plate capacitor has a dielectric with constant 3.5. The plates are circular with radius 0.07 m, and the separation gap is 0.375 mm. What energy may be stored in this capacitor if 1250 V is the applied voltage?

Question

Quizplus · Accepted Answer

The energy stored in a capacitor can be calculated using the formula:

$U = \frac{1}{2}CV^2$

where U is the energy stored, C is the capacitance, and V is the voltage.

The capacitance of a parallel plate capacitor with dielectric K and separation distance d can be calculated using the formula:

$C = \frac{K\epsilon_0A}{d}$

where $\epsilon_0$ is the permittivity of free space and A is the area of the plates.

Substituting the given values, we get:

$A = \pi r^2 = \pi (0.07	ext{ m})^2 = 0.0154	ext{ m}^2$

$d = 0.375	ext{ mm} = 0.000375	ext{ m}$

$K = 3.5$

$\epsilon_0 = 8.85 	imes 10^{-12}	ext{ F/m}$

$C = \frac{K\epsilon_0A}{d} = \frac{(3.5)(8.85 	imes 10^{-12}	ext{ F/m})(0.0154	ext{ m}^2)}{0.000375	ext{ m}} = 2.263 	imes 10^{-11}	ext{ F}$

Now, we can use the formula for energy to find the answer:

$U = \frac{1}{2}CV^2 = \frac{1}{2}(2.263 	imes 10^{-11}	ext{ F})(1250	ext{ V})^2 = 9.9 	imes 10^{-4}	ext{ J} = \boxed{9.9 	imes 10^{-4}	ext{ J}}$

A Parallel Plate Capacitor Has a Dielectric with Constant 3