Question

A parallel plate capacitor of capacitance 90pF is connected to a battery of emf 20V. If a dielectric material of K=$\dfrac{5}{3}$ is inserted between the plates, then find the magnitude of the induced charge?
(a) 2.5nC
(b) 0.9nC
(c) 1.2nC
(d) 0.3nC

Answer 1

The capacitance tells us how much charge the device stores for a given voltage. When a dielectric is introduced inside the capacitor. Then, capacitance increases by some factor of dielectric constant i.e. K.
So Q=CV (without any dielectric between the plates of parallel plates of the capacitor).
And ${C_{new}} = KC$ (after inserting a dielectric between parallel plates it's original capacitance is altered by a 'K' (dielectric constant) factor)
Due to the increase in capacitance of the conductor, it would store more charge than before.

Formula Used:
1.Total charge stored by capacitor:${Q_{new}} = {C_{new}}V$
2. Induced charge in the capacitor: ${Q_{in}} = {Q_{new}} - Q$ …… (A)
3. $1F = {10^{12}}pF = {10^9}nF$ (conversion of 1 Farad(F) into Pico farad(pF) and nano farad(nF))

Complete step by step solution:
Given:
Emf across capacitor: V=20 V
Capacitance of capacitor: C=90 pF
Magnitude of dielectric constant: K=$\dfrac{5}{3}$
Step 1 of 4:
The total charge stored in the capacitor without insertion of a dielectric

Q = CV \\\ \Rightarrow Q = 90 \times {10^{ - 12}} \times 20C \\\ \Rightarrow Q = 1.8 \times {10^{ - 9}}C = 1.8nC \\\ $$ …… (1) Step 2 of 4: When dielectric inserted new capacitance becomes $K$ times: ${C_{new}} = KC$ $ \Rightarrow {C_{new}} = KC = \dfrac{5}{3} \times 90pF$ $ \Rightarrow {C_{new}} = 150pF$ Step 3 of 4: The total charge stored when the dielectric is inserted:(using the value of new capacitance from step 2) ${Q_{new}} = {C_{new}}V = 150pF \times 20V$ ${Q_{new}} = 3nC$ …… eq (2) Step 4 of 4: Using equation (2) and (1) and substituting in equation (A) we get total charge induced: ${Q_{induced}} = (3 - 1.8)nC = 1.2nC$ **$\therefore$ The magnitude of the induced charge is ${Q_{induced}} = (3 - 1.8)nC = 1.2nC$. So, Option (C) is correct.** **Additional information:** Here, only one dielectric material is inserted in the entire region of space between parallel plates. But the same problem can be made complicated by the partitioning region between plates. For such problems, we need to use Kirchhoff’s voltage sum law. Capacitors are widely used as backup supplies in the AC circuit around the load in order to minimize the sudden voltage drop across the load which is in general very expensive. Sudden voltage change can result in a short-circuit of the load. **Note:** The introduction of dielectric material inside the capacitor induces an electric field inside the dielectric material. This Electric field makes the charges inside to rearrange itself to get more charge on the surface of parallel plate capacitors.