Answer:
The value is [tex]V_d = 5 \ V[/tex]
Explanation:
From the question we are told that
The capacitance of the capacitor is [tex]C = 0.16 \ pF = 0.16*10^{-12} \ F[/tex]
The voltage is [tex]V = 10\ V[/tex]
The number of electrons present on the negative plate is [tex]N_e = 1.00 *10^{7} \ electrons[/tex]
Generally the charge on the positive plate at 10 Volt is mathematically represented as
[tex]Q_a = C * V[/tex]
=> [tex]Q_a = 0.16*10^{-12} * 10[/tex]
=> [tex]Q_a = 0.16*10^{-11} \ C[/tex]
Now the charge on the plate when the electron where placed is evaluated as
[tex]Q_b = Q_a + ( N_e * e)[/tex]
Where e is the charge on each electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]Q_b = 0.16*10^{-11} + (1.0*10^7 * 1.60*10^{-19})[/tex]
=> [tex]Q_b = 3.2 *10^{-12 } \ C[/tex]
Generally the voltage between the two plate is evaluated as
[tex]V_d = \frac{ Q_b - Q_a }{ 2 * C }[/tex]
=> [tex]V_d = \frac{ 3.2*10^{-12} - 0.16*10^{-11}}{2 * 0.16 *10^{-12}}[/tex]
=> [tex]V_d = 5 \ V[/tex]
