Chapter 33, Problem 1P

(a)

To determine

The time rate of increase of electric field between the plates.

Answer to Problem 1P

The time rate of increase of electric field between the plates is $7.19\times {10}^{11}\text{V}/\text{m}\cdot \text{s}$.

Explanation of Solution

Given info: The current is $0.200\text{A}$, the radius of the circular plate is $10.0\text{cm}$ and the separation distance between the plates is $4.00\text{mm}$.

The value of permittivity of free space is $8.85\times {10}^{-12}\text{F}/\text{m}$ and the value of the permeability constant is $\text{4π}\times {10}^{-7}\text{H}/\text{m}$

The area of the circular plate is,

$A=\text{π}{r}^2$

Here,

$r$ is the radius of the circular plate.

The formula for the capacitance of the capacitor is.

$\begin{array}{c}C=\frac{A{\epsilon }_0}{d}\\ =\frac{\text{π}{r}^2{\epsilon }_0}{d}\end{array}$

Here,

${\epsilon }_0$ is the permittivity of free space.

$d$ is the separation between the plates.

Substitute $10.0\text{cm}$ for $r$, $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ and $4.00\text{mm}$ for $d$ in above equation to find $C$.

$\begin{array}{c}C=\frac{\text{π}{\left(10.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}{4.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =6.95\times {10}^{-11}\text{F}\end{array}$

Thus, the capacitance of the capacitor is $6.95\times {10}^{-11}\text{F}$.

The formula for the capacitance of the capacitor is.

$\begin{array}{c}C=\frac{Q}{V}\\ V=\frac{Q}{C}\end{array}$

Here,

$Q$ is the charge of the plates.

$V$ is the potential of the plates.

Differentiate the above equation with respect to time.

$\frac{dV}{dt}=\frac{\frac{dQ}{dt}}{C}$

The formula for the current is,

$I=\frac{dQ}{dt}$

Then,

$\frac{dV}{dt}=\frac{I}{C}$

The formula for the electric field is,

$E=\frac{V}{d}$

Differentiate the above equation with respect to time.

$\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}$

Substitute $\frac{I}{C}$ for $\frac{dV}{dt}$ in above equation.

$\frac{dE}{dt}=\frac{I}{Cd}$

Substitute $0.200\text{A}$ for $I$, $6.95\times {10}^{-11}\text{F}$ for $C$ and $4.00\text{mm}$ for $d$ in above equation to find $\frac{dE}{dt}$.

$\begin{array}{c}\frac{dE}{dt}=\frac{0.200\text{A}}{\left(6.95\times {10}^{-11}\text{F}\right)\left(4.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}\\ =7.19\times {10}^{11}\text{V}/\text{m}\cdot \text{s}\end{array}$

Conclusion:

Therefore, the time rate of increase of electric field between the plates is $7.19\times {10}^{11}\text{V}/\text{m}\cdot \text{s}$.

(b)

To determine

The magnetic field between the plates $5.00\text{cm}$ from the centre.

Answer to Problem 1P

The magnetic field between the plates $5.00\text{cm}$ from the centre is $2.0\times {10}^{-7}\text{T}$.

Explanation of Solution

Given info: The current is $0.200\text{A}$, the radius of the circular plate is $10.0\text{cm}$ and the separation distance between the plates is $4.00\text{mm}$.

The formula for the electric flux through the circular plate is,

$\begin{array}{c}{\Phi }_{\text{E}}=EA\\ \frac{d{\Phi }_{\text{E}}}{dt}=A\frac{dE}{dt}\\ =\text{π}{r}^2\frac{dE}{dt}\end{array}$

The Maxwell’s equation is.

$\begin{array}{c}{\displaystyle \oint \overrightarrow{\text{B}}\cdot \text{d}\overrightarrow{\text{s}}}={\epsilon }_0{\mu }_0\frac{d{\Phi }_{\text{E}}}{dt}\\ B\left(2\text{π}r\right)={\epsilon }_0{\mu }_0\text{π}{r}^2\frac{dE}{dt}\\ B=\left(\frac{r{\epsilon }_0{\mu }_0}{2}\right)\frac{dE}{dt}\end{array}$

Substitute $5.00\text{cm}$ for $r$, $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$, $\text{4π}\times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_0$ and $7.19\times {10}^{11}\text{V}/\text{m}\cdot \text{s}$ for $\frac{dE}{dt}$ in above equation to find $B$.

$\begin{array}{c}B=\left(\frac{\left(5.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)\left(\text{4π}\times {10}^{-7}\text{H}/\text{m}\right)}{2}\right)\left(7.19\times {10}^{11}\text{V}/\text{m}\cdot \text{s}\right)\\ =1.9990\times {10}^{-7}\text{T}\\ \cong 2.0\times {10}^{-7}\text{T}\end{array}$

Conclusion:

Therefore, the magnetic field between the plates $5.00\text{cm}$ from the centre is $2.0\times {10}^{-7}\text{T}$.