Answer:
The potential difference is [tex]V = 8.5 *10^{-2} \ volt[/tex]
Explanation:
From the question we are told that
The diameter of the of both plates is [tex]D = 2.50 *10^{-3} \ m[/tex]
The distance of separation is [tex]d = 1.40 *10^{-4} \ m[/tex]
The potential difference is [tex]V = 0.12 \ V[/tex]
The new charge on the plate is [tex]q = 70.7[/tex]% of [tex]Q_o[/tex]
Where [tex]Q_o[/tex] is the original charge on the capacitor
Generally the charge on the capacitor is mathematically represented as
[tex]Q = CV[/tex]
Where C is the capacitance of the capacitor which is mathematically represented as
[tex]C = \epsilon_o \frac{A}{d}[/tex]
Where [tex]\epsilon_o[/tex] is the permittivity of free space which is a constant with value
[tex]\epsilon_o = 8.85 *10^{-12} F/m[/tex]
A is the area which is mathematically represented as
[tex]A = \pi \frac{D^2}{4}[/tex]
substituting values
[tex]A = 3.142 * \frac{6.25*10^{-6}}{4}[/tex]
[tex]A = 4.909*10^{-6} \ m^2[/tex]
Now
[tex]C = 8.85 *10^{-12} \frac{4.909 *10^{-6}}{1.40 *10^{-4}}[/tex]
[tex]C = 3.1014 *10^{-13} \ F[/tex]
So
[tex]Q_o = 3.1014 *10^{-13}* 0.12[/tex]
[tex]Q_o = 3.72 *10^{-14} \ C[/tex]
Generally
[tex]q= CV[/tex]
Here
[tex]q = \frac{70.7}{100} * Q_o[/tex]
[tex]q = \frac{70.7}{100} * 3.72*10^{-14}[/tex]
[tex]q = 2.636*10^{-14} \ C[/tex]
Making V the subject in the above formula
[tex]V = \frac{q}{C}[/tex]
Substituting values
[tex]V = \frac{2.636 *10^{-14}}{3.1014 *10^{-13} }[/tex]
[tex]V = 8.5 *10^{-2} \ volt[/tex]
