Question

1/KThe charge on an isolated capacitor does not change when a sheet of glass is inserted between the capacitor plates, and we find that the potential difference decreases (because the electric field inside the insulator is reduced by a factor of $\mathbf{1}\mathbf{/}\mathit{K}$ ). Suppose instead that the capacitor is connected to a battery, so that the battery tries to maintain a fixed potential difference across the capacitor. (a) A light bulb and an air-gap capacitor of capacitance$\mathit{C}$ are connected in series to a battery with known emf. What is the final charge$\mathit{Q}$ on the positive plate of the capacitor? (b) After fully charging the capacitor, a sheet of plastic whose dielectric constant$\mathit{K}$ is inserted into the capacitor and fills the gap. Does any current run through the light bulb? Why? What is the final charge on the positive plate of the capacitor?

Short answer

The final charge of positive plate of the capacitor is the product of the capacitance and emf of the battery.

Step-by-step solution

Write the given data from the question.

The battery tries to main the fix potential difference across the capacitor,

$\mathit{E}\mathit{m}\mathit{f}\mathbf{=}\mathit{\Delta }\mathit{V}$

The dielectric constant of the plastic sheet is $\mathit{K}$.

The electric field inside the capacitor reduce by the factor$\mathbf{1}\mathbf{/}\mathit{K}$.

Determine the formulas to calculate final charge on the positive plate of capacitor.

The potential difference between the plates of the capacitor is defined as the ratio of the charge and capacitance of the capacitor.

The expression to calculate the potential difference of the capacitor is given as follows.

$\mathit{\Delta }\mathit{V}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{C}}$ …… (i)

Here,$\mathit{Q}$ is the charge, and$\mathit{C}$ is the capacitance of capacitor.

Calculate final charge on the positive plate of capacitor.

Calculate the charge of the capacitor.

$\mathit{Q}\mathbf{=}\mathit{C}\mathit{\Delta }\mathit{V}$

Substitute$\mathit{E}\mathit{m}\mathit{f}$ for$\mathit{\Delta }\mathit{V}$ into above equation.

$\mathit{Q}\mathbf{=}\mathit{C}\left(Emf\right)$

The final charge of the capacitance is direction proportional to the emf of the battery.

Hence, the final charge of positive plate of the capacitor is the product of the capacitance and emf of the battery.