Question

Two parallel plates of area 2.34.10-³ m²

have 7.07.10-7 C of charge placed on
them. A 6.62.10-5 C charge q₁ is placed
between the plates. What is the
magnitude of the electric force on q₁?
Hint: How is force related to the field?
[?] N
Submit

Answer 1

The electric force F on a charge q in an electric field E is given by the equation
`\(F = q \cdot E\)`. The electric field between the plates E can be determined using the formula
`\(E = (V)/(d)\)`, where V is the potential difference between the plates and d is the separation between the plates.

First, calculate the electric field:

`\[ E = (V)/(d) \]`

The potential difference V between the plates can be calculated using the formula
`\(V = Ed\)`, where A is the area of the plates, Q is the charge on the plates, and
`\(\varepsilon_0\)` is the permittivity of free space:

`\[ V = Ed = (Q)/(\varepsilon_0 A) \]`

Substitute this value of E into the force equation:

`\[ F = q \cdot E = q \cdot (V)/(d) \]`

Now, substitute the known values and calculate the magnitude of the electric force F.