Question

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 increase the plate separation, do the following quantities increase, decrease, or stay the same?

Answer 1

The capacitance characteristics allow to find that the response to increase the separation of the plates are:

• Capacitance increases

• The potential difference decreases

• The charge remains constant

A capacitor is a system formed by two parallel plates charged with opposite charge that stores energy, the capacitance is

            C =     $\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}$  

Where C is the capacitance, Q the stored charge, ΔV the voltage difference, A the area of ​​the plates, d the separation between them and ε₀ the dielectric permittivity (ε₀ = 8.85 10⁻¹² $\frac{C^2}{N \ m^2 }$)

In this case the capacitor is charged and then the separation of the plates increases, the different amounts are affected:

•  From the expression it is observed that the capacitance increases

•  The area of ​​the plates remains constant,

•  The charge remains constant, since it is not connected to the battery

•  The potential difference decreases.

In conclusion of the characteristics of capacitance, let us find that the response for increased separation of the plates are:

• Capacitance increases

• The potential difference decreases

• The charge remains constant

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