Question

(a) Find the expression for the induced charge on a surface of a linear dielectric slab (dielectric constant $K$) inserted between the parallel plates of a charged capacitor fully occupying the intervening region.

(b) Four capacitors, each of capacitance $1\,\mu\text{F}$ are to be connected to obtain a net capacitance of $0.75\,\mu\text{F}$. Draw a diagram to show the required combination.

Hint:

Net field in the slab is the plate field minus the induced field, and it equals E0/K. For 0.75 microfarad, use three in parallel with one in series.

Answer 1

Step 1: Understanding the Concept:
A dielectric slab placed in an electric field gets polarised. Its atoms form tiny dipoles, and bound charges of opposite signs appear on the two faces.
These induced charges produce a field inside the slab that opposes the applied field. So the net field inside becomes smaller.

Step 2: Key Formula or Approach:
Let the plates carry free charge density \(\pm\sigma\) and let the induced surface density on the slab faces be \(\mp\sigma_p\).
Field without the slab: \(E_0 = \sigma/\varepsilon_0\). Field with the slab: \(E = E_0/K\), by the definition of dielectric constant.

Step 3: Part (a): Induced charge:
Inside the slab the net field is the plate field minus the field of the induced charges:
\[ E = E_0 - E_p = \frac{\sigma}{\varepsilon_0} - \frac{\sigma_p}{\varepsilon_0} \]
Put \(E = E_0/K = \sigma/(K\varepsilon_0)\):
\[ \frac{\sigma}{K\varepsilon_0} = \frac{\sigma - \sigma_p}{\varepsilon_0} \Rightarrow \sigma_p = \sigma\left(1 - \frac{1}{K}\right) \]
Multiply by the plate area A. With free charge \(Q = \sigma A\), the induced charge is:
\[ Q_p = Q\left(1 - \frac{1}{K}\right) = \frac{Q(K-1)}{K} \]

Step 4: Part (b): Choose the combination:
Parallel connection adds capacitances and series connection lowers them. Target: \(0.75\ \mu F\), which is less than \(1\ \mu F\), so a series link is needed.
Join three capacitors in parallel: \(C_p = 1+1+1 = 3\ \mu F\). Connect this block in series with the fourth capacitor of \(1\ \mu F\).
\[ C_{eq} = \frac{3 \times 1}{3 + 1} = \frac{3}{4} = 0.75\ \mu F \]
Diagram in words: one battery terminal leads to a junction where three 1 \(\mu F\) capacitors sit side by side (parallel). Their other ends meet at a second junction. From there one 1 \(\mu F\) capacitor connects in series to the other battery terminal.

Final Answer:
The induced charge is \(Q_p = Q(1 - 1/K)\). Three capacitors in parallel, in series with the fourth, give 0.75 \(\mu F\). \[ \boxed{Q_p = Q\left(1-\frac{1}{K}\right),\quad C_{eq}=0.75\ \mu F} \]