Question

Two charges of unknown magnitudes are kept 5m away from each other. The magnitude of force between them is measured to be F. If these charges are then moved so that the distance between them is 20m, what is the new force between them?

• A. $\frac{F}{16}$
• B. $\frac{F}{4}$
• C. $\frac{F}{8}$
• D. $\frac{F}{20}$

Answer 1

The correct option is A

$\frac{F}{16}$

(a) Ok, how do you approach this question without knowing charge values? Well first try solving it directly

$F={\frac{K{q}_1{q}_2}{5}}^2=\frac{K{q}_1{q}_2}{25}$

In the second case,

$F_2=\frac{K{q}_1{q}_2}{{20}^2}=\frac{K{q}_1{q}_2}{400}$

$\frac{F_2}{F}=\frac{25}{400}=\frac{1}{16}$

$F_2=\frac{F}{16}$

Or in a simpler way,

$F\propto \frac{1}{d^2}$

Proportionalities can be converted to ratios (think about why)

$\frac{F_1}{F_2}=\frac{d_2^2}{d_1^2}$

$\frac{F_1}{F_2}=\frac{(4\times 5{)}^2}{5^2}$

$\frac{F_1}{F_2}=16$

$F_2=\frac{F_1}{16}$