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5 March, 18:31

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

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Answers (1)
  1. 5 March, 22:07
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    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.
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