Question

The force between two charges when they are 2 cm apart is
0.036 N. If the sum of two charges is 10uC, what are the
charges? (1/4ttɛo=9x109 Nm-C-2).​

Answer 1

Answer:

$q_1=9.9998\mu C$ and $q_2=0.0002\mu C$

Or

$q_1=0.00016\mu C$ and $q_2=9.99984\mu C$

Explanation:

We are given that

Force between two charges=0.036 N=$36\times 10^{-3}N$

Distance between two charges, r=2cm=$2\times 10^{-2}$m

1m=100cm

Sum of two charges=$10\mu C$

Let one charge=$q_1=q\mu C=q\times 10^{-6}C$

$q_2=(10-q)\times 10^{-6} C$

We know that

Electric force between two charges

$F=\frac{kq_1q_2}{r^2}$

Where $k=\frac{1}{4\pi \epsilon_0}=9\times 10^{9}$

Using the formula

$36\times 10^{-3}=9\times 10^{9}\times \frac{q\times 10^{-6}\times(10-q)\times 10^{-6}}{(2\times 10^{-2})^2}$

$\frac{144\times 10^{-7}}{9\times 10^{9}\times 10^{-12}}=q(10-q)$

$0.0016=10q-q^2$

$q^2-10q+0.0016=0$

$10000q^2-100000q+16=0$

$q=\frac{100000\pm\sqrt{(100000)^2-4\times 10000\times 16}}{2\times 10000}$

Using the formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$q=9.999$ and $q=0.00016$

$q_2=10-9.9998=0.0002$

$q_2=10-0.00016=9.99984$

Hence, two charges are

$q_1=9.9998\mu C$ and $q_2=0.0002\mu C$

Or

$q_1=0.00016\mu C$ and $q_2=9.99984\mu C$