A parallel plate capacitor is constructed using two square metal sheets,each of side L = 10 cm.The plates are separated by a distance d = 2 mm and a voltage applied between the plates.The electric field strength within the plates is E = 4000 V/m.The energy stored in the capacitor is
A)0.71 nJ
B)1.42 nJ
C)2.83 nJ
D)3.67 nJ
E)zero

Question

Quizplus · Accepted Answer

The energy stored in a capacitor can be calculated using the formula $U = \frac{1}{2} \epsilon_0 E^2 A d$, where $U$ is the energy, $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $E$ is the electric field strength, $A$ is the area of the plates, and $d$ is the separation between the plates. Given $E = 4000 \, \text{V/m}$, $L = 10 \, \text{cm} = 0.1 \, \text{m}$, and $d = 2 \, \text{mm} = 0.002 \, \text{m}$, the area $A = L^2 = (0.1)^2 = 0.01 \, \text{m}^2$. Plugging these values into the formula gives $U = \frac{1}{2} (8.85 \times 10^{-12}) (4000)^2 (0.01) (0.002) = 1.42 \times 10^{-9} \, \text{J} = 1.42 \, \text{nJ}$.

A Parallel Plate Capacitor Is Constructed Using Two Square Metal