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Chapter 25: Q. 8 (page 709)

Two positive point charges are $5.0 cm$ apart. If the electric potential energy is $72 渭闯$ , what is the magnitude of the force between the two charges?

Short Answer

Expert verified

The force between the two charges has a magnitude of $1.4 脳 10^{- 3} N$ .

Step by step solution

01

Formula for magnitude of force

Given information:

Distance between two positive charges is $5.0 c m$ .

Electric potential energy is $72 渭 J$ .

Find:

The objective is to find magnitude of the force between two charges.

The potential energy is calculated as follows:

$U = \frac{1}{4 {蟺蔚}_{0}} \frac{Q_{1} Q_{2}}{r} . . 1$

when the force's magnitude is

$F = \frac{1}{4 {蟺蔚}_{0}} \frac{Q_{1} Q_{2}}{r^{2}} . . . . . . . 2$

$Q_{1}$ -Charge of point $1$

$Q_{2}$ -Charge of point $2$

$r$ -Distance

$U$ -Electric potential energy

02

Calculation for magnitude of force

when the force's magnitude is

$F = \frac{1}{4 {蟺蔚}_{0}} \frac{Q_{1} Q_{2}}{r^{2}}$

Equation $2$ divided by equation $1$ ,

we obtain,

$\frac{F}{U} = \frac{\frac{1}{r^{2}}}{\frac{1}{r}} = \frac{1}{r}$

This results in

$F = \frac{1}{r} U$

$= \frac{72 渭闯}{5.0 cm}$

localid="1649412963669" $= 1.4 脳 10^{- 3} N$

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