# Chapter 22 - Electric Charges and Forces - Exercises and Problems - Page 626: 58

The force is increasing at a rate of $~~7.2\times 10^{-4}~N/s$

#### Work Step by Step

We can find the rate at which the force is increasing: $F = \frac{kq^2}{r^2}$ $\frac{dF}{dt} = \frac{2kq}{r^2}~\frac{dq}{dt}$ $\frac{dF}{dt} = \frac{(2)(9.0\times 10^9~N~m^2/C^2)(5.0\times 10^{-9}~C)}{(0.025~m)^2}~\cdot (5.0\times 10^{-9}~C/s)$ $\frac{dF}{dt} = 7.2\times 10^{-4}~N/s$ The force is increasing at a rate of $~~7.2\times 10^{-4}~N/s$

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