Question

What will be the effect on capacitance of capacitor, when dielectric substance is inserted between the plates of the capacitor?

Hint:

The dielectric constant $\kappa$ is also called relative permittivity ($\varepsilon_r$). It represents how many times the capacitance increases compared to vacuum. Higher $\kappa$ means better charge storage capacity.

Answer 1

Definition of Capacitance: 

Capacitance ($C$) is the ability of a capacitor to store electric charge. It is defined as the ratio of the charge ($Q$) stored on each plate to the potential difference ($V$) between the plates: 

$$ C = \frac{Q}{V} $$ 
For a parallel plate capacitor, the capacitance is given by: 

$$ C = \frac{\varepsilon_0 A}{d} $$ 
where:

• $\varepsilon_0$ = Permittivity of free space ($8.85 \times 10^{-12} \, F/m$)
• $A$ = Area of each plate
• $d$ = Distance between the plates

Effect of Inserting a Dielectric: 

When a dielectric substance (insulating material) is inserted between the plates of a capacitor, the capacitance increases by a factor of $\kappa$ (dielectric constant or relative permittivity). 

$$ C_{\text{with dielectric}} = \kappa \, C_{\text{without dielectric}} = \frac{\kappa \varepsilon_0 A}{d} $$ 
where $\kappa > 1$ for all dielectric materials. 

Why Does Capacitance Increase? 

1. Reduction of Electric Field:

• When a dielectric is inserted, its molecules become polarized.
• Positive charges align toward the negative plate and negative charges align toward the positive plate.
• This creates an induced electric field ($E_{\text{induced}}$) that opposes the original electric field ($E_0$).

Net electric field: 

$$ E_{\text{net}} = E_0 - E_{\text{induced}} = \frac{E_0}{\kappa} $$ 

2. Decrease in Potential Difference: 
Since for a uniform electric field: 

$$ V = E \cdot d $$ 
A decrease in electric field causes a decrease in potential difference: 

$$ V_{\text{with dielectric}} = \frac{V_0}{\kappa} $$ 

3. Increase in Capacitance: 
From: 

$$ C = \frac{Q}{V} $$ 
If charge $Q$ remains constant and $V$ decreases, then $C$ increases. 

If the capacitor remains connected to a battery (constant $V$), more charge $Q$ flows onto the plates, again increasing capacitance. 

Two Important Cases: 

Case 1: Battery Disconnected (Constant Charge)

• Charge remains constant: $Q = \text{constant}$
• Potential difference decreases: $V = \frac{V_0}{\kappa}$
• Capacitance increases: $C = \kappa C_0$
• Energy stored decreases:

$$ U = \frac{Q^2}{2C} $$ 

Case 2: Battery Connected (Constant Voltage)

• Voltage remains constant: $V = \text{constant}$
• Charge increases: $Q = \kappa Q_0$
• Capacitance increases: $C = \kappa C_0$
• Energy stored increases:

$$ U = \frac{1}{2} C V^2 $$ 

Energy Stored with Dielectric: 

$$ U = \frac{1}{2} C V^2 = \frac{1}{2} \kappa C_0 V^2 $$ 

Dielectric Constants of Common Materials: 

 

Material	Dielectric Constant ($\kappa$)
Vacuum	1.0000
Air	1.0006
Paper	3.7
Glass	5 – 10
Mica	5 – 7
Rubber	7
Distilled Water	80
Barium Titanate	1000 – 10000

Additional Effects of Dielectric:

• Increased Breakdown Voltage: Dielectrics allow capacitors to operate at higher voltages without electrical breakdown.
• Mechanical Support: Provides physical separation between capacitor plates.
• Enhanced Energy Storage: More energy can be stored due to increased capacitance.

Final Result: 

Capacitance increases by a factor of dielectric constant when a dielectric is inserted: 

$$ C_{\text{with dielectric}} = \kappa \, C_{\text{without dielectric}} $$