Question

Two equal point charges 2.5 cm apart, both initially neutral, are being charged at the rate of 5.0 nC/s. At what rate (N/s) is the force between them increasing 1.0 s after charging begins?

Short answer

The rate at which the force is increasing is$7.12\times {10}^{-4}N/s$.

Step-by-step solution

Given Information

The given point charges are equal

Separation of the two points$=r=2.5cm=2.5\times {10}^{-2}m$

Rate of change of charge$=\frac{dq}{dt}=5.0nC/s=5.0\times {10}^{-9}C/s$

Calculation

Since it is given that the charges are the same, the electrostatic force can be given as:

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

Where,

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}=8.9\times {10}^{9}N{m}^2/C^2$

By differentiating the above equation with respect to time, we get the rate of change of force as:

$\frac{dF}{dt}=\frac{1}{4\pi {\epsilon }_0{r}^2}2q\frac{dq}{dt}.....\left(1\right)$

The charge of the particles after 1.0 s can be calculated as:

localid="1648461224815" $$\begin{gathered} q=\frac{dq}{dt}t \\ q=5.0\times {10}^{-9}\times 1.0 \\ q=5.0\times {10}^{-9}C \end{gathered}$$

By substituting this value and the given values in equation (1), we get,

$$\begin{gathered} \frac{dF}{dt}=\frac{\left(8.9\times {10}^9\right)}{{\displaystyle {\left(2.5\times {10}^{-2}\right)}^2}}\times 2\times \left(5.0\times {10}^{-9}\right)\left(5.0\times {10}^{-9}\right) \\ \frac{{\displaystyle dF}}{{\displaystyle dt}}=7.12\times {10}^{-4}N/s \end{gathered}$$

Final answer

Hence, for the given conditions, the force is increasing at a rate of$7.12\times {10}^{-4}N/s$.