An empty plate capacitor is connected between the terminals ofa 9.0 V battery and charged up. The capacitor is then disconnectedfrom the battery, and the spacing between the capacitor plates isdoubled. As a result of the change, what is the new voltage betweenthe plates of the capacitor?
I have:
E = v/d
E =9 V
then the distance is doubled 9(2) = 18 V
I feel like maybe I over simplified the problem. Does thislook right?

The new voltage between the parallel plates of the capacitor is 18V, because for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage) between the plates.

Explanation:

ΔV = E*Δd

Where;

ΔV is the change in potential difference

Δd is the change in the distance between the parallel plates

E is the electric field potential.

Assuming a constant electric field; [tex]E = \frac{v}{d} , then; \frac{v_1}{d_1} =\frac{v_2}{d_2}[/tex]

when the spacing between the capacitor plates is doubled, d₂ = 2d₁

v₂ = (v₁*d₂)/(d₁)

v₂ = (v₁*2d₁)/(d₁)

v₂ = 2v₁

v₂ = 2(9) = 18 V

Therefore, for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage).