Question

Compute the force of attraction between a +1.60 x 10-19 C charge and a – 2.09 x 10-18 C charge if they are 4 x10-10 m apart.

Answer 1

Since the Electrostatic force=( K X q1q2)/d^2

=>9.1 X 10^-31 X 1.6 X 10^-19 X -2.09 X 10^-18/(4 X10^-10)^2

=> -1.9019 * 10^-48 N

Answer 2

The force of attraction between the two charges will be 1.881 × 10⁻⁸ N

• As the two charges are opposite in nature, i.e., one positive and one negative, so the nature of the force will be attractive.

The value of the charges are,

q₁ = + 1.60 × 10⁻¹⁹ C

q₂ = - 2.09 × 10⁻¹⁸ C        

Distance between them (r) = 4 × 10⁻10 m

According to the law of electrostatic force,

F = $\frac{1}{4\pi\epsilon } \frac{q_{1} q_{2} }{r^2}$  ε₀= permittivity of free space

or, F = (9 × 10⁹ × 1.60 × 10⁻¹⁹ × 2.09 × 10⁻¹⁸) / (4 × 10⁻10)² N  [Value of 1/4π∈ is

                                                                                                    9 × 10⁹ N m²/C²]

or, F = 1.881 × 10⁻⁸ N

So, the force of attraction between the two charges is 1.881 × 10⁻⁸ N .