Question

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the plates. What is the magnitude of the electric force on q1?

Answer 1

Answer: 2250 N

Step-by-step explanation:

Answer 2

The force exerted on the
`q_1` is
`F = 2.25*10^(3) \ N`

Step-by-step explanation:

From the question we are told that

The area is
`A = 2.34*10^(-3) \ m^2`

The magnitude of charge placed on them is
`q = 7.07 * 10^(-7) C`

The charge placed between the plate is
`q_1 = 6.62 *10^(-5) C`

The electric field generated around the plate is mathematically represented as

`E = (q)/(A \epsilon_o)`

Substituting values

`E = (7.07*10^(-7))/(2.34*10^(-3) * 8.85 *10^(-12))`

`E = 34*10^(6) \ V/m`

The force exerted the charge
`q_1` is mathematically represented as

`F = q_1 * E`

Substituting values

`F = 6.62 *10^(-5) * 34*10^(6)`

`F = 2.25*10^(3) \ N`