Question

A capacitor with a slab of glass between the plates is connected to a battery by Nichrome wires and allowed to charge completely. Then the slab of glass is removed. Describe and explain what happens. Include diagrams. If you give a direction for a current, state whether you are describing electron current or conventional current.

Short answer

The electric field reduces, and capacitance increases, due to which charge storage decreases.

Step-by-step solution

Write the given data from the question.

The slab of the glass is placed between the plates of the capacitor.

After charging the completely, the glass slab removed from the capacitor.

Determine the formulas to calculate the capacitance of the capacitor.

The expression to calculate the capacitance of the capacitor when glass slab is placed between the plates is given as follows.

$\mathit{C}\mathbf{=}\mathit{K}\frac{\mathbf{A}{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{d}}$

Here, $\mathit{K}$is the dielectric constant of material, $\mathit{d}$is the distance between the plates,$\mathit{A}$ area of the plates and ${\mathit{\epsilon }}_{\mathbf{0}}$.

Determine the effect of removing the glass on the charge.

When the glass slab is placed between the plates, the electric field due to the glass slab is in the opposite direction of the electric field due to the plates. The net electric field is reduced, and capacitance increases, which increases stored charge.

The value for the air,$K_{air}=1$ , and for the glass,$K_{glass}>1$, by this we can see capacitance is increasing.

Therefore, the direction of the electric field and conventional current when the glass slab is placed between the plates are shown below.

If the glass slab is removed between the capacitor plates, the net electric field increases and capacitance decreases, due to which the charge storage decreases.

Therefore, the direction of the electric field and conventional current when the glass slab is removed between the plates are shown below.