Question

two equal and opposite charges
of magnetude 2× 10^8 C are
15 com apart.
Find the force between them.
​

Answer 1

Given,

q1 = q2 = q = $2*10^{8}C$

Distance between the charges = 15 cm = $15*10^{-2}m$

To find,

Force between them = $\frac{1}{4\pi\epsilon_0}\frac{q^{2}}{r^{2}}$

                                    = $9*10^{9}*\frac{(2*10^{8})^{2}}{(15*10^{-2)^{2}}}$

                                    = $9*10^{9}*\frac{4*10^{16}}{225*10^{-4}}$

                                    = $0.16*10^{24} N$

More information :

If two point charges q1 and q2  are situated at a distance r apart, then the force acting between them is given by ,

F ∝ $\frac{q1q2}{r^{2}}$$F = k \frac{q1q2}{r^{2}}$

k = proportionality constant.