Answer:
The magnitude of the charge on each electrode is 77.8 nC.
Explanation:
Given that,
Length = 4.0 cm
Distance = 2.0 mm
Electric field strength [tex]E=5.5\times10^{6}\ N/C[/tex]
We need to calculate the magnitude of the charge on each electrode
Using formula of electric field
[tex]E=\dfrac{\sigma}{\epsilon_{0}}[/tex]
Where, [tex]\sigma[/tex] = charge density
We know that,
The charge density is
[tex]\sigma=\dfrac{Q}{A}[/tex]
Put the value of charge density into the formula of electric field
[tex]E=\dfrac{Q}{A\epsilon_{0}}[/tex]
[tex]Q=EA\epsilon_{0}[/tex]
Put the value into the formula
[tex]Q=5.5\times10^{6}\times4.0\times4.0\times10^{-4}\times8.85\times10^{-12}[/tex]
[tex]Q=77.8\times10^{-9}\ C[/tex]
[tex]Q=77.8\ nC[/tex]
Hence, The magnitude of the charge on each electrode is 77.8 nC.