Question

The force between two charges is 4 × 10^–9 N . If the magnitude of one charge is reduced by a factor of two and the distance between the charges is reduced by a factor of two, what is the new force between the charges?

A. 2 × 10^–9 N
B. 4 × 10^–9 N
C. 6 × 10^–9 N
D. 8 × 10^–9 N

Answer 1

`8 * 10^(-9)\; {\rm N}`.

Step-by-step explanation:

By Coulomb's Law, the magnitude of electrostatic force between two point charges is:

`\displaystyle F = (k\, q_(1)\, q_(2))/(r^(2))`,

Where:

• 
  `k` is Coulomb's Constant,
• 
  `q_(1)` and
  `q_(2)` are the magnitudes of the two charges, and
• 
  `r` is the distance between the two charges.

In this question, assume that the magnitude of the two point charges were originally
`q_(1)` and
`q_(2)` with a distance of
`r` in between.

Assume that
`q_(2)` becomes
`(q_(2) / 2)` and
`r` becomes
`(r / 2)`. By Coulomb's Law, the magnitude of the electrostatic force between the two new charges would become:

`\begin{aligned}F &= (k\, q_(1)\, (q_(2) / 2))/((r / 2)^(2)) \\ &= (k\, q_(1)\, q_(2) / 2)/(r^(2) / 2^(2)) \\ &= (2\, k\, q_(1)\, q_(2))/(r^(2))\end{aligned}`.

In other words, magnitude of the force between the two new charges would be twice that of the original value. The magnitude of the new force would be
`8 * 10^(-9)\; {\rm N}`.