Answer:
[tex]\displaystyle F^{\prime}(x) = \frac{-1.8 \times 10^{9}\, Q \, q}{x^{3}}\; \rm N\cdot m^{-1}[/tex] assuming that the units of [tex]Q[/tex] and [tex]q[/tex] are both coulomb.
If [tex]Q = q = 2\; \rm C[/tex], then [tex]F^{\prime}(0.001\; \rm m) = -7.20\times 10^{19}\; \rm N \cdot m^{-1}[/tex].
Explanation:
It is given that:
[tex]\displaystyle F(x) = \frac{k \cdot Q \cdot q}{x^{2}}[/tex].
Equivalently:
[tex]\displaystyle F(x) = k \cdot Q \cdot q \cdot x^{-2}[/tex].
Differentiate [tex]F(x)[/tex] with respect to [tex]x[/tex] to find the corresponding instantaneous rate of change.
By the power rule for derivatives, [tex]\begin{aligned} \frac{d}{dx}\, \left[x^{-2}\right] &= (-2)\, x^{-3} \end{aligned}[/tex].
Likewise, when differentiating [tex]F(x)[/tex]:
[tex]\begin{aligned}F^{\prime}(x) &= \frac{d}{dx}\left[k\cdot Q \cdot q \cdot x^{-2}\right] \\ &= k \cdot Q \cdot q \cdot \frac{d}{dx} \left[x^{-2}\right] \\ &= (-2)\, k \cdot Q \cdot q \cdot x^{-3} \\ &= \frac{(-2)\, k \cdot Q \cdot q}{x^{3}}\end{aligned}[/tex].
It is given that [tex]k = 9 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex]. Assume that the units of [tex]Q[/tex] and [tex]q[/tex] are both coulomb, and that the unit of [tex]x[/tex] is meter (such that no unit conversion would be necessary,) substitute the value of [tex]k[/tex] into the expression for [tex]F^{\prime}(x)[/tex]:
[tex]\begin{aligned} F^{\prime}(x) &= \frac{(-2)\, k \cdot Q \cdot q}{x^{3}} \\ &= \frac{(-2) \times 9 \times 10^{9}\; {\rm N \cdot m^{2} \cdot C^{-2}} \cdot (Q\; {\rm C}) \cdot (q \; {\rm C})}{({x\;\rm m})^{3}} \\ &= \frac{1.8 \times 10^{10}\, Q \cdot q}{x^{3}}\; \rm N \cdot m^{-1}\end{aligned}[/tex].
Substitute in [tex]Q = 2[/tex], [tex]q = 2[/tex], and [tex]x = 0.001[/tex] into the expression and evaluate:
[tex]\begin{aligned}F^{\prime}({0.001\; \rm m}) &= \frac{1.8 \times 10^{10}\times 2 \times 2}{0.001^{3}}\; \rm N \cdot m^{-1} \\ &= 7.20 \times 10^{19}\; \rm N \cdot m^{-1} \end{aligned}[/tex].
