A 7.5 $\mu$ F parallel-plate capacitor having a plate separation distance of 0.25 mm has 8.2 mC of charge (magnitude per plate). The electric field between the plates is
A) $8.6 \times 10 ^ { 3 }$ N/C.
B) $4.4 \times 10 ^ { 6 }$ N/C.
C) $7.5 \times 10 ^ { 4 }$ N/C.
D) $1.6 \times 10 ^ { 5 }$ N/C.

Question

Quizplus · Accepted Answer

The electric field $E$ in a parallel-plate capacitor is given by $E = \frac{V}{d}$, where $V$ is the voltage across the capacitor and $d$ is the separation between the plates. The voltage $V$ can be found from the charge $Q$ and capacitance $C$ using the formula $V = \frac{Q}{C}$. Substituting $Q = 8.2 \times 10^{-3}$ C (since 1 mC = $10^{-3}$ C) and $C = 7.5 \times 10^{-6}$ F into $V = \frac{Q}{C}$ gives $V = \frac{8.2 \times 10^{-3}}{7.5 \times 10^{-6}} = 1093.33$ V. Then, using $d = 0.25 \times 10^{-3}$ m, the electric field $E = \frac{1093.33}{0.25 \times 10^{-3}} = 4.37332 \times 10^{6}$ N/C, which is closest to choice B.

A 75 μ\muμ F Parallel-Plate Capacitor Having a Plate Separation Distance of 0

A 75 $\mu$ F Parallel-Plate Capacitor Having a Plate Separation Distance of 0