You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. when you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same?
a. c
b. q
c. e between the plates
d. delta-v

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.

b

explanation:

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s