Question

Two point charges of 8.4×10−6 C, on the left, and −5.4×10−6 C, directly to its right, are separated by 0.024 m. Find the electrostatic force acting on the charge on the left. A force to the right is taken as positive, while one directed to the left is negative.

Answer 1

Solution :

The magnitude of Coulomb force acting between two charged particles is given by

$|F|=\frac{k|q_1||q_2|}{r^2}$

Where

q₁ = 8.4 $\times$ 10^-6 C

q₂ = -5.4 $\times$ 10^-6

r = distance between charges = 0.024 m

k = constant = 9 $\times$ 10⁹ N.m²/C²

putting all the values

$|F|=\frac{9\times 10^9 \times 8.4 \times 10^{-6}\times 5.4 \times 10^{-6}}{0.024^2} = 708.750\ N$

Now as the charge on right is negative so it will apply an attractive force on the charge on left that the direction of above force will be towards right, hence the force on left charge is

$F = 708.750\ N$.................Answer