Three point charges $$q$$, $$-2q$$ and $$2q$$ are placed on x axis at a distance $$x = 0$$, $$x = \frac{3}{4}R$$ and $$x = R$$ respectively from origin as shown. If $$q = 2 \times 10^{-6}$$ C and $$R = 2$$ cm, the magnitude of net force experienced by the charge $$-2q$$ is _____ N.

Given charges:

• +q at x=0
• $$−2q\ at\ x=\frac{3R}{4}​\ (this\ is\ the\ ch\arg e\ on\ which\ force\ is\ required)\ $$

• $$+2q\ at\ x=R$$

Given:

$$q=2\times10^{-6}C$$

$$R=2cm=0.02m$$

Force on -2q due to charge q

Distance between them:

$$r_1=\frac{3R}{4}$$

Coulomb force:

$$F_1=\frac{k(q)(2q)}{(3R/4)^2}$$

$$\frac{2kq^2}{9R^2/16}$$

$$\frac{32kq^2}{9R^2}$$

Using

$$k=9\times10^9$$

and

$$q^2=(2\times10^{-6})^2=4\times10^{-12}$$

$$R^2=(0.02)^2=4\times10^{-4}$$

$$F_1=\frac{32(9\times10^9)(4\times10^{-12})}{9(4\times10^{-4})}=320N$$

This force is toward left (attraction).

Force on -2q due to charge 2q

Distance:

$$r_2=R-\frac{3R}{4}=\frac{R}{4}$$

Force:

$$F_2=\frac{k(2q)(2q)}{(R/4)^2}$$

$$=\frac{4kq^2}{R^2/16}$$

$$=\frac{64kq^2}{R^2}$$

Substituting:

$$F_2=\frac{64(9\times10^9)(4\times10^{-12})}{4\times10^{-4}}=5760N$$

This is toward right (attraction).

Net force on -2q is

Opposite directions, so subtract:

$$F=5760-320=5440N$$