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

Step-by-step 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=(kq_1q_2)/(d^2)`

`F_1=(9* 10^9* 4* 2* 10^(-14))/((5* 10^(-2))^2)`

`F_1=28.8\ N`

`F_2=(kq_3q_2)/(d^2)`

`F_2=(9* 10^9* 6* 2* 10^(-14))/((5* 10^(-2))^2)`

`F_2=43.2\ N`

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