A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectr

Q: Why does the free charge on the plates change when the dielectric is inserted?

Because the battery remains connected, it forces the voltage $V$ to stay constant. Inserting a dielectric increases the capacitance $C$. Since $Q = CV$, an increase in $C$ with a constant $V$ means the charge $Q$ must also increase. The battery supplies this extra charge.

Q: What exactly is 'induced charge'?

When a dielectric is placed in an electric field, its molecules polarize (align with the field). This alignment causes a net charge to appear on the surfaces of the dielectric. This is called the induced or bound charge, and it creates an electric field that opposes the original field.

Q: Where does the formula $Q_{ind} = Q(1 - 1/K)$ come from?

It comes from Gauss's Law applied to dielectrics. The net electric field inside the dielectric is $E_{net} = E_0 / K$. Since $E_{net}$ is also produced by the net charge $(Q - Q_{ind})$, we have $(Q - Q_{ind}) / (A\epsilon_0) = (Q / A\epsilon_0) / K$. Solving for $Q_{ind}$ gives the formula.

Q: What would happen if the battery was disconnected before inserting the dielectric?

If the battery was disconnected, the free charge $Q$ on the plates would be trapped and remain constant at its initial value $Q_0 = CV_0$. The induced charge would then simply be $Q_0(1 - 1/K)$, and the voltage would drop to $V_0/K$.

Q: Does the thickness of the dielectric matter for the induced charge?

If the dielectric fully fills the space between the plates, its thickness doesn't change the formula for the induced charge on its surface. The induced charge depends only on the free charge $Q$ and the dielectric constant $K$.

Question

A parallel plate capacitor of capacitance $90	ext{ pF}$ is connected to a battery of emf $20	ext{ V}$. If a dielectric material of dielectric constant $K=\frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be

Options:

(A) $1.2	ext{ nC}$

(B) $0.3	ext{ nC}$

(C) $2.4	ext{ nC}$

(D) $0.9	ext{ nC}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a parallel plate capacitor connected to a battery. The battery acts as a constant voltage source, maintaining a potential difference of $V = 20	ext{ V}$ across the plates. Initially, the capacitance is $C = 90	ext{ pF}$.

When a dielectric material with a dielectric constant $K = \frac{5}{3}$ is inserted between the plates, the capacitance of the system increases to $C' = KC$. Because the battery remains connected, the voltage $V$ cannot change. To maintain this voltage with a higher capacitance, the battery must pump more charge onto the plates. The new free charge on the plates becomes:

$$Q = C'V = KCV$$

### The Physics of Induced Charge

Now, let's look at what happens inside the dielectric. The electric field created by the free charges on the plates polarizes the molecules within the dielectric material. This polarization results in a net bound charge, or **induced charge** ($Q_{ind}$), appearing on the surfaces of the dielectric facing the plates.

The magnitude of this induced charge is always a fraction of the free charge $Q$ on the plates, given by the fundamental relation:

$$Q_{ind} = Q \left(1 - \frac{1}{K}ight)$$

### The Master Equation

Since we know that the new free charge is $Q = KCV$, we can substitute this into our induced charge formula:

$$Q_{ind} = KCV \left(1 - \frac{1}{K}ight)$$

By distributing the $K$ inside the parenthesis, the equation simplifies beautifully:

$$Q_{ind} = CV(K - 1)$$

This elegant formula tells us exactly how much charge is induced based purely on the initial parameters and the dielectric constant!

### Final Calculation

We have all the necessary values to find the answer. Let's plug them in carefully:

$$Q_{ind} = (90 	imes 10^{-12}	ext{ F}) 	imes (20	ext{ V}) 	imes \left(\frac{5}{3} - 1ight)$$

First, calculate the initial charge $CV$:

$$90 	imes 20 = 1800$$

Next, evaluate the term in the bracket:

$$\frac{5}{3} - 1 = \frac{2}{3}$$

Now, multiply them together:

$$Q_{ind} = 1800 	imes 10^{-12} 	imes \frac{2}{3} = 1200 	imes 10^{-12}	ext{ C}$$

To express this in a more standard unit, we convert picocoulombs to nanocoulombs:

$$Q_{ind} = 1.2 	imes 10^{-9}	ext{ C} = \mathbf{1.2	ext{ nC}}$$

This perfectly matches option (a). Always remember to check whether the battery is connected or disconnected, as it completely changes the behavior of the free charge!

Animated Solution for Physics - Electrostatics: A parallel plate capacitor of capacitance $90 pF$ is connected to a battery of emf $20 V$ . If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be

Select Answer:

Visualized Solution

Initial Setup

Dielectric Insertion

New Charge on Plates

Induced Charge Formula

Expressing Induced Charge

Substituting Values

Calculation

Final Answer

The Way Forward

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Physics of Induced Charge

The Master Equation

Final Calculation

Similar Questions

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 1 0^{4}$ V/m, the charge density of the positive plate will be close to

A parallel plate capacitor is made of two square plates of side ' $a$ ' separated by a distance $d$ ( $d ≪ a$ ). The lower triangular portions is filled with a dielectric of dielectric constant $K$ , as shown in the figure. Capacitance of this capacitor is

A parallel plate capacitor with plates of area $1 m^{2}$ each, are at a separation of $0.1 m$ . If the electric field between the plates is $100 N/C$ , the magnitude of charge on each plate is $(Take, ε_{0} = 8.85 \times 1 0^{- 12} \frac{C ^{2}}{N-m ^{2}})$

A parallel plate capacitor has $1 μ F$ capacitance. One of its two plates is given $+ 2 μ C$ charge and the other plate $+ 4 μ C$ charge. The potential difference developed across the capacitor is