Question

A particle with a net charge of -2 μC is halfway between two particles with a net charge of +4 μC and +6 μC, respectively. The two positively charged particles are 10 cm apart. What is the magnitude of the net force on the negative particle?A) 72 NB) 14.4 NC) 3.6 ND) 18 N

Answer 1

Answer:B

Explanation:

Given

Two charged particle with charge $4\mu C$ and $6 \mu C$

distance between them $r=10 cm$

Third particle of $-2 \mu C$ is situated between the two

Force on negative charge due to the 6 \mu C charge

$F_1=\frac{kq_1q_2}{d^2}$

$F_1=\frac{9\times 10^9\times 4\times 2\times 10^{-14}}{(5\times 10^{-2})^2}$

$F_1=28.8\ N$

$F_2=\frac{kq_3q_2}{d^2}$

$F_2=\frac{9\times 10^9\times 6\times 2\times 10^{-14}}{(5\times 10^{-2})^2}$

$F_2=43.2\ N$

$F_2-F_1=43.2-28.8=14.4\ N$