Question

In the diagram, q1 = +6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? Include a + or - sign to indicate the direction.

Answer 1

822.202925

Step-by-step explanation:

I swear

Answer 2

E =+823.12N/C.

Step-by-step explanation:

The electric field
`E` at point
`P` is the sum of electric field
`E_1` due to
`q_1` and
`E_2` due to
`q_2` at
`P`:

`E = E_1+E_2`.

The distance from
`P` to
`q_1` is
`R_1 = 0.150m+0.250m = 0.400m`; therefore, the electric due to it is

`E_1 = k(q_1)/(R_1^2)`

`E_1 = 9*10^9*(+6.39*10^(-9)C)/((0.400m)^2)`

`E_1 = +359.44N/C.`

And the distance from
`P` to
`q_2` is
`R_2 = 0.250m`; therefore,

`E_2 = k(q_2)/(R_2^2)`

`E_2 = 9*10^9*(+3.22*10^(-9)C)/((0.250m)^2)`

`E_2 = +463.68N/C`

Hence, the total electric field at point
`P` is

`E = E_1+E_2`

`E = +359.44N/C+463.68N/C`

`\boxed{E =+823.12N/C.}`

where the plus sign indicates that the the electric field points to the right.