Answer:
The charge must be on each plate is [tex]1.881\times10^{-6}\ C[/tex].
Explanation:
Given that,
Electric field [tex]E= 8.5\times10^{5}\ V/m[/tex]
Area = 2500 cm²
Distance = 0.10 mm
We need to calculate the potential difference
Using formula of potential difference
[tex]\Delta V=E\times d[/tex]
[tex]\Delta V=8.5\times10^{5}\times0.10\times10^{-3}[/tex]
[tex]\Delta V=85\ V[/tex]
We need to calculate the capacitor
Using formula of capacitor
[tex]C=\dfrac{\epsilon_{0}\times A}{d}[/tex]
Put the value into the formula
[tex]C=\dfrac{8.85\times10^{-12}\times2500\times10^{-4}}{0.10\times10^{-3}}[/tex]
[tex]C=2.2125\times10^{-8}\ F[/tex]
We need to calculate the charge
Using formula of charge
[tex]Q=C\times\Delta V[/tex]
[tex]Q=2.2125\times10^{-8}\times85[/tex]
[tex]Q=1.881\times10^{-6}\ C[/tex]
Hence, The charge must be on each plate is [tex]1.881\times10^{-6}\ C[/tex].
