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

Answer:

Charge on A is $q=0.7820\times 10^{-5}C$

Charge on B is $2q=2\times 0.7820\times 10^{-5}C=1.5640\times 10^{-5}C[/tex] Explanation:We have given one charge is twice of other charge Let [tex]q_1=q$, 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

$F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}$, here K is constant which value is $9\times 10^9$

So $43=\frac{9\times 10^92q^2}{0.16^2}$

$q^2=0.0611\times 10^{-9}$

$q^2=0.611\times 10^{-10}$

$q=0.7820\times 10^{-5}C$ so charge on A is $q=0.7820\times 10^{-5}C$

And charge on B is $2q=2\times 0.7820\times 10^{-5}C=1.5640\times 10^{-5}C$