Respuesta :
Answer:
a. C will decrease
b. Q will remain the same
c. E will decrease
d. Delta-V will increase
Explanation:
Justification for C:
As we know that for parallel plate capacitors, capacitance is calculated using:
C = (ϵ_r * ϵ_o * A) / d - Say it Equation 1
Where:
ϵ_r - is the permittivity of the dielectric material between two plates
ϵ_o - Electric Constant
A - Area of capacitor's plates
d - distance between capacitor plates
From equation 1 it is clear that capacitance will decrease if distance between the plates will be increased.
Justification of Q
As charge will not be able to travel across the plates, therefore it will remain the same
Justification of E
As we know that E = Delta-V / Delta-d, thus considering Delta-V is increasing on increasing Delta-d (As justified below) as both of these are directly proportional to each other, therefore Electric field (E) will remain constant as capacitors' plates are being separated.
Moreover, as the E depends on charge density which remains same while plates of capacitor are being separated therefore E will remain the same.
Justification of Delta-V
As we know that Q = C * V, therefore considering charge remains the same on increasing distance between plates, voltage must increase to satisfy the equation.
