Question

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on the x axis, and a 70-nc charge at y = –6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?

Answer 1

Answer:48 V

Step-by-step explanation:

Given

Three charged particle with charge

`q_1=50\ nC at y=6\ m`

`q_2=-80\ nC at x=-4\ m`

`q_3=70\ nC at y=-6\ m`

Electric Potential is given by

`V=(kQ)/(r)`

Distance of
`q_1` from
`x=8\ m`

`d_1=√(6^2+8^2)`

`d_1=√(36+64)`

`d_1=10\ m`

similarly
`d_2=8-(-4)`

`d_2=12\ m`

`d_3=√((-6)^2+8^2)`

`d_3=√(36+64)`

`d_3=10\ m`

Potential at
`x=8\ m` is

`V_(net)=(kq_1)/(d_1)+(kq_2)/(d_2)+(kq_3)/(d_3)`

`V_(net)=k[(q_1)/(d_1)+(q_2)/(d_2)+(q_3)/(d_3)]`

`V_(net)=9* 10^9[(50)/(10)-(80)/(12)+(70)/(10)]* 10^(-9)`

`V_(net)=9* 5.33`

`V_(net)=47.97\approx 48\ V`