Worked example: Electric force between plates
Two parallel metal plates are separated by $3.5 \mathrm{~cm}$ and have a potential difference of $7.9 \mathrm{kV}$. Calculate the electric force acting on a stationary charged particle between the plates that has a charge of $2.6 \times 10^{-15} \mathrm{C}$.
Answer/Explanation
Step 1:
Write down the known values
Potential difference, $\Delta V=7.9 \mathrm{kV}=7.9 \times 10^3 \mathrm{~V}$
Distance between plates, $\Delta \mathrm{d}=3.5 \mathrm{~cm}=3.5 \times 10^{-2} \mathrm{~m}$
Charge, $Q=2.6 \times 10^{-15} \mathrm{C}$
Step 2:
Calculate the electric field strength between the parallel plates
$
E=\frac{\Delta V}{\Delta d}
$
$
E=\frac{7.9 \times 10^3}{3.5 \times 10^{-2}}=2.257 \times 10^5 \mathrm{~V} \mathrm{~m}^{-1}
$
Step 3:
Write out the equation for electric force on a charged particle
$
\mathbf{F}=\mathbf{Q E}
$
Step 4: Substitute electric field strength and charge into electric force equation
$
F=Q E=\left(2.6 \times 10^{-15}\right) \times\left(2.257 \times 10^5\right)=5.87 \times 10^{-10} \mathrm{~N}=5.9 \times 10^{-10} \mathrm{~N}(2 \text { s.f. })
$