Respuesta :
Answer:
[tex]V_1=8 V_2[/tex]
Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2, [tex]d_2=d[/tex]
- separation distance of capacitor 1, [tex]d_1=2d[/tex]
- quantity of charge on capacitor 2, [tex]Q_2=Q[/tex]
- quantity of charge on capacitor 1, [tex]Q_1=4Q[/tex]
We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
[tex]C=\frac{k.\epsilon_0.A}{d}[/tex].....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air
[tex]\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}[/tex]
From eq. (1)
For capacitor 2:
[tex]C_2=\frac{k.\epsilon_0.A}{d}[/tex]
For capacitor 1:
[tex]C_1=\frac{k.\epsilon_0.A}{2d}[/tex]
[tex]C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}][/tex]
We know, potential differences across a capacitor is given by:
[tex]V=\frac{Q}{C}[/tex]..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
[tex]V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}[/tex]
[tex]V_2=\frac{Q.d}{k.\epsilon_0.A}[/tex]
& for capacitor 1:
[tex]V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}[/tex]
[tex]V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}[/tex]
[tex]V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}][/tex]
[tex]V_1=8 V_2[/tex]
