Question

a parallel - plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm. if the charge on each plate has a magnitude of 3 x 10⁻⁶ c, then the force exerted by one plate on the other has a magnitude of about
o 8 x 10⁶ n
o 5 n
o 1 x 10⁴ n
o 2.5 n
o 9 x 10⁵ n

Explanation:

Step1: Recall the formula for force between plates of a capacitor

$F=\frac{Q^{2}}{2\epsilon_0 A}$
where $Q$ is the charge on the plate, $\epsilon_0 = 8.85\times 10^{- 12}\ C^{2}/N\cdot m^{2}$ is the permittivity of free - space and $A$ is the area of the plate.
Given $Q = 3\times10^{-6}\ C$, $A=0.2\ m^{2}$.

Step2: Substitute the values into the formula

$F=\frac{(3\times 10^{-6})^{2}}{2\times8.85\times 10^{-12}\times0.2}$
First, calculate $(3\times 10^{-6})^{2}=9\times 10^{-12}$.
Then, the denominator is $2\times8.85\times 10^{-12}\times0.2 = 3.54\times 10^{-12}$.
So, $F=\frac{9\times 10^{-12}}{3.54\times 10^{-12}}\approx 2.5\ N$

Answer:

$2.5\ N$