Question

Two point charges, A and B, are separated by a distance of 16.0cm. The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 43.0 N on the other, find the magnitudes of the charges. Charge A: ____ in C

Charge B: _____ in C

Answer 1

Charge on A is
`q=0.7820* 10^(-5)C`

Charge on B is
, then
`q_2=2q`

Distance between two charges = 16 cm = 0.16 m

Force F = 43 N

According to coulombs law force between tow charges is given by

`F=(1)/(4\pi \epsilon _0)(q_1q_2)/(r^2)=(Kq_1q_2)/(r^2)`, here K is constant which value is
`9* 10^9`

So
`43=(9* 10^92q^2)/(0.16^2)`

`q^2=0.0611* 10^(-9)`

`q^2=0.611* 10^(-10)`

`q=0.7820* 10^(-5)C` so charge on A is
`q=0.7820* 10^(-5)C`

And charge on B is
`2q=2* 0.7820* 10^(-5)C=1.5640* 10^(-5)C`