Question

when the distance between charged parallel plates of a capacitor is d, the potential difference is v. if the distance is decreased to d/2, how will the potential difference change, if at all? the new potential difference would be four times the previous value. the new potential difference would be one - fourth of the previous value. the new potential difference would be one - half of the previous value. the new potential difference would be the same as the previous value. the new potential difference would be twice the previous value.

Explanation:

Step1: Recall electric - field and potential - difference relation

The potential difference $V$ between the plates of a parallel - plate capacitor is given by $V = Ed$, where $E$ is the electric field between the plates and $d$ is the distance between the plates. Assuming the electric field $E$ remains constant (for a capacitor with a fixed charge, $E=\frac{\sigma}{\epsilon_0}$, and $\sigma$ is the surface - charge density which is constant when the charge on the capacitor is constant).

Step2: Analyze the change in potential difference

Let the initial distance be $d_1 = d$ and the initial potential difference be $V_1=Ed_1$. The new distance is $d_2=\frac{d}{2}$. Then the new potential difference $V_2 = Ed_2$. Substituting $d_2=\frac{d}{2}$ into the formula, we get $V_2 = E\times\frac{d}{2}$. Since $V_1 = Ed$, we can see that $V_2=\frac{1}{2}V_1$.

Answer:

The new potential difference would be one - half of the previous value.