Chapter 22, Problem 25P

(a)

To determine

The energy delivered in each pulse.

Answer to Problem 25P

The answer is $280J$ .

Explanation of Solution

Given data:

Power, $P=2.8\times {10}^{11}W$

Time, $t=2.5\times {10}^{-7}s$

Formula used:

It is known that: $U=Pt$

Where,

  $\text{U}$ is the energy emitted in each pulse.

  $P$ is the power.

  $t$ is the time.

Calculation:

Substitute the given data into the formula $U=Pt$ ,

  $U=Pt=2.8\times {10}^{11}W\times 2.5\times {10}^{-7}s=280J$

Conclusion: The energy delivered in each pulse is $280J$ .

(b)

To determine

The value of root mean square for the electric field will be determined.

Answer to Problem 25P

The answer is $2.63\times {10}^9\text{V/m}$ .

Explanation of Solution

Given data:

Power, $P=2.8\times {10}^{11}W$

Time, $t=2.5\times {10}^{-7}s$

Radius, $r=2.2\times {10}^{-3}m$

Speed of the light, $c=3\times {10}^8\text{m/s}$

Permittivity of vacuum, ${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}$

Formula used:

It is known that: $S=\frac{P}{A}=c{\epsilon }_0{E}_{rms}{}^2$

Where,

  $\text{S}$ is the power flux.

  $P$ is the power.

  $c$ is speed of the light.

  $E_{rms}$ is root square means value of the electric field.

  ${\epsilon }_0$ is the permittivity of vacuum.

Calculation:

Substitute the given data into the formula $S=\frac{P}{A}$ ,

  $S=\frac{P}{A}=\frac{2.8\times {10}^{11}W}{\pi {\left(2.2\times {10}^{-3}m\right)}^2}=1.8415\times {10}^{16}W/m^2$

Now, we calculate the root mean square value of the electric field,

  $S=c{\epsilon }_0{E}_{rms}{}^2=1.8415\times {10}^{16}W/m^2=\left(3\times {10}^8\text{m/s}\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)E_{rms}{}^2$

Simplify further,

  $E_{rms}=\sqrt{\frac{1.8415\times {10}^{16}W/m^2}{\left(3\times {10}^8\text{m/s}\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)}}=2.63\times {10}^9\text{V/m}$

Conclusion: The root mean square value of the electric field is $2.63\times {10}^9\text{V/m}$ .