Question

Two point charges are 3 m apart and their combined charge is 8 uc. the force of repulsion between them is 0.012 n. charges are:

Answer 1

Answer: 6 and 2 microC

Explanation:

Answer 2

The value of charges are 2 and 6.

Given,

The combined charge is = 8 μC

i.e., the sum of the charges is = 8 μC

Suppose, one charge is = q μC

                                        = q × 10⁻⁶ C

∴ The other charge will be = (8 - q) μC

                                            = (8 - q) × 10⁻⁶ C

The force of repulsion between them = 0.012 N

The distance between them (r) is = 3 m

As the force is repulsive, so the two charges have same sign.

According to the formula of Electrostatic force,

   F = $\frac{1}{4\pi\epsilon_{0} } \frac{q_{1}q_{2} }{r^2} }$

∴ 0.012 = 9 × 10⁹ $\frac{q(8-q)}{3^2}$        [$\frac{1}{4\pi\epsilon_{0} }$ = 9 × 10⁹  S.I. unit]

⇒ 0.012/ 10⁹ = 8q - q²

By solving this, we get,

The value of charges are 2 and 6.