Question

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

What is the magnitude of the electrical force between the two charges:
answer to the closest 0.1 N

Answer 1

Approximately
`5.5\; \rm N`, assuming that the volume of these two charged objects is negligible.

Step-by-step 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
`q_1` and
`q_2` denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question,
`q_1 = 20 * 10^(-6)\; \rm C` and
`q_2 = 15 * 10^(-6)\; \rm C`.

Let
`r` denote the distance between these two point charges. In this question,
`r = 0.7\; \rm m`.

Let
`k` denote the Coulomb constant. In standard units,
`k \approx 8.98755* 10^(9)\; \rm kg \cdot m^(3)\cdot s^(-2)\cdot C^(-2)`.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

`\begin{aligned}F &= (k \cdot q_1 \cdot q_2)/(r^(2))\end{aligned}`.

Substitute in the values and evaluate:

`\begin{aligned}F &= (k \cdot q_1 \cdot q_2)/(r^(2))\\ &\approx 8.98755 * 10^(9)\; \rm kg \cdot m^(3)\cdot s^(-2)\cdot C^(-2) \\ &\quad * 20* 10^(-6)\; \rm C\\ &\quad * 15* 10^(-6)\; \rm C \\ &\quad * \frac{1}{{(0.7\; \rm m)}^(2)}\\ &\approx 5.5\; \rm N \end{aligned}`.