Question

You make repeated measurements of the electric field$\overrightarrow{\mathbf{E}}$due to a distant charge, and you find it is constant in magnitude and direction. At timerole="math" localid="1656916621351" $\mathbf{t}\mathbf{=}\mathbf{0}$your partner moves the charge. The electric field doesn’t change for a while, but at time$\mathbf{t}\mathbf{=}\mathbf{45}\mathbf{}\mathbf{ns}$you observe a sudden change. How far away was the charge originally?

Short answer

Originally the charge was at a distance of 13.5 m .

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The time at which the charge moved by the partner is,$t=0$
• The time at which the change was observed is, $t=45\mathrm{ns}\times \left(\frac{{10}^{-9}\mathrm{s}}{1\mathrm{ns}}\right)=45\times {10}^{-9}\mathrm{s}$.

Significance of the distance

The distance moved by an object is equal to the product of the velocity of that object and the time taken by that object.

The equation of the distance of the charge is expressed as:

$d=vt$ …(i)

Here, $d$is the distance moved by the charge, $v$is the velocity of the chargethat is the speed of ,the light and $t$is the time taken.

Determination of the distance of the charge

The concept of retardation is being applied here as the change of the velocity of the electric field is equal to the distance of the speed of light.

As the change has been observed after light took about to reach charge to the distance. However, the light travels about 1 ft in every 1 ns hence the charge happens at the distance of 45 ft .

Substitute $3\times {10}^8\mathrm{m}/\mathrm{s}$for $v$and$45\times {10}^{-9}\mathrm{s}$ for $t$in equation (i).

$$\begin{gathered} d=\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\times \left(45\times {10}^{-9}\mathrm{s}\right) \\ =13.5\mathrm{m} \end{gathered}$$

Thus, the charge was at a distance of $13.5\mathrm{m}$.