Question

In the diagram, q1 = +6.39*10^_9 C and

q2 = +3.22*10^-9 C. What is the
electric potential at point P?
Include a + or - sign.

Answer 1

259.695

Step-by-step explanation:

Acellus

Answer 2

The electric potential will be "259.695 volt".

Step-by-step explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

`q_1=6.39* 10^(-9) \ C`

`q_2=3.22* 10^(-9) \ C`

`AP=(0.150+0.250)`

`=0.40 \ m`

`BP=0.25 \ m`

Now,

At point P, the electric potential will be:

⇒
`V=(q_1)/(4 \pi \epsilon_o AP ) +(q_2)/(4 \pi \epsilon_o BP)`

By putting values, we get

⇒
`=9* 10^9 [(6.39* 10^(-9))/(0.40) +(3.22* 10^(-9))/(0.25) ]`

⇒
`=259.695 \ Volt`