Chapter 17, Problem 39P

To determine

The voltage across the capacitor.

Answer to Problem 39P

The answer is $4.5\times {10}^{4}V/m$ .

Explanation of Solution

Given:

Charges $Q=72\mu C=72\times {10}^{-6}\text{}C$

Capacitance is, $0.80\mu F=0.80\times {10}^{-6}F$

Distance between the plates is $2.0mm=0.002m$

Formula used:

It is also known that $C={\in }_0\frac{A}{d}$ , here $C$ is capacitance, $A$ is area, $d$ is distance and ${\in }_0$ is permittivity of free space.

  ${\in }_0=8.85\times {10}^{-12}{C}^2/N\cdot m^2$

Calculation:

The voltage across capacitor is magnitude of the electric field times separation distance of plates.

By using the expression $Q=CV=CEd\to E=\frac{Q}{Cd}$

Substitute all the given value:

  $=\frac{\left(72\times {10}^{-6}C\right)}{\left(0.80\times {10}^{-6}F\right)\left(2.0\times {10}^{-3}m\right)}=4.5\times {10}^{4}V/m$

Conclusion:

The required value is $4.5\times {10}^{4}V/m$ .