Answer:
Approximately [tex]5.5\; \rm N[/tex], assuming that the volume of these two charged objects is negligible.
Explanation:
Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.
Let [tex]q_1[/tex] and [tex]q_2[/tex] denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, [tex]q_1 = 20 \times 10^{-6}\; \rm C[/tex] and [tex]q_2 = 15 \times 10^{-6}\; \rm C[/tex].
Let [tex]r[/tex] denote the distance between these two point charges. In this question, [tex]r = 0.7\; \rm m[/tex].
Let [tex]k[/tex] denote the Coulomb constant. In standard units, [tex]k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}[/tex].
By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:
[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}[/tex].
Substitute in the values and evaluate:
[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}[/tex].
