Parallel Plate Capacitor Questions

Parallel plate capacitors are essential components in understanding electrostatics and energy storage. These questions cover key principles such as capacitance, electric fields, dielectric effects, and potential difference, helping build a solid foundation in physics. Parallel plate capacitors are widely used in electronics and physics applications. These questions help you understand how they function, how their capacitance changes with geometry and materials, and how to apply formulas effectively.

1.0Concept of Capacitors

• A capacitor or condenser consists of two conductors separated by an insulator or dielectric. Having equal and opposite charges on which sufficient quantity of charge may be accommodated.
• It is a device which is used to store energy in the form of an electric field by storing charge. Conductors are used to form capacitors.

2.0Electrical Capacitance

• It demonstrates a conductor's ability to store electrical energy through an electric field. If charge(Q) is given to an isolated conducting body and its potential increases by V, then $Q\propto V,Q=CV$

$\Rightarrow C=\frac{Q}{V}$ (C = Capacitance of the capacitor)

• Electrical capacitance is a Scalar quantity.
• Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor.

3.0Graph Between Q and V

• S I Unit- Farad
• CGS Unit-Stat Farad
• Dimensional Formula-[M^-1L^-2T⁴A²]

4.0Circuit Symbols of Capacitor

5.0Capacitors Applications

Capacitors find extensive applications across diverse fields because of their efficient capability to store and discharge electrical energy.

1. Filtering and Smoothing
2. Energy Storage
3. Timing Circuits
4. Motor Starters
5. Power Factor Correction
6. Memory Backup

6.0Parallel Plate Capacitor(PPC)

• It consists of two large plane parallel conducting plates separated by small distance
• $C=\frac{{ϵ}_{0}A}{d}$
• Capacitance of parallel plate capacitor with dielectric, $C=\frac{{ϵ}_r{ϵ}_{0}A}{d}={ϵ}_r{C}_0(\because C_0=\frac{{ϵ}_{0}A}{d})$

Capacitance of Parallel Plate Capacitor Depends on

1. Area C ∝ A
2. Distance between the plates $\Rightarrow C\propto \frac{1}{d}$
3. Medium between the Plates $\Rightarrow C\propto {ϵ}_r$

7.0Capacitance of Parallel Plate Capacitor when Dielectric is Partially Filled

Capacitance is given by,

$C=\frac{{ϵ}_{0}A}{(d-t)+\frac{t}{{ϵ}_r}}$

8.0FAQ on Parallel Plate Capacitor Questions

Q-1.A small parallel plate capacitor has its plates charged as shown. A particle with charge q is placed at a distance l from the capacitor (where l≫d, the separation between the plates). What is the magnitude of the force acting on the charged particle due to the capacitor?

Solution:

The two plates acts as a dipole

The magnitude of force on charge q 

$\vec{F}=|\vec{E}|q=\left(\frac{2kqd}{r^3}\right)q=\frac{qad}{2\pi {ϵ}_0{l}^3}$

Q-2.A parallel plate capacitor with plate area A and plate distance d is charged to a potential difference V, after which the battery is disconnected. What is the work done in increasing the separation between the plates to 2d?

Solution: As the battery is disconnected, charge remains constant in the work process.

Work done = final potential energy – initial potential energy

$\text{Work done}=\frac{Q^2}{2C^{\prime }}-\frac{Q^2}{2C^0}=\frac{Q^2}{2}\left(\frac{1}{C^{\prime }}-\frac{1}{C^0}\right)$

$$\begin{gathered} \text{Where}Q=CV=\frac{A{ϵ}_{0}V}{d}‘C=\frac{A{ϵ}_0}{d},C^{\prime }=\frac{A{ϵ}_0}{2d} \\ \text{Now,work done}=\frac{{ϵ}_0{V}^{2}A}{2d} \end{gathered}$$

Q-3.Two capacitors of capacitance 1 F and 2Fare charged to potential difference 20Vand 15V as shown in figure. If now terminal B and C are connected together terminal A with positive battery and D with negative terminal of battery 30 V then find out final charges on both the capacitors.

Solution:

Now applying Kirchoff voltage law,

$-\frac{20+q}{1}-\frac{30+q}{2}+30=0$

$-40-2q-30-q=-60$

$3q=-10\Rightarrow q=-\frac{10}{3}\mu C$

$\text{Charge flow}=-\frac{10}{3},\mu C$

$Chargeoncapacitorofcapacitance1\mu F$$=20+q=\frac{50}{3}\mu C$

$Chargeoncapacitorofcapacitance2\mu F$$=30+q=\frac{80}{3}\mu C$

Q-4.Find out equivalent capacitance between A and B. (take each plate Area = A)

Solution:

Let charge density as shown

$C_{eq}=\frac{Q}{V}=\frac{2xA}{V}$

$V=V_2-V_4=(V_2-V_3)+(V_3-V_4)=\frac{xd}{{ϵ}_0}+\frac{2xd}{{ϵ}_0}=\frac{3xd}{{ϵ}_0}$

$C_{eq}=\frac{Q}{V}=\frac{2xA{ϵ}_0}{3xd}=\frac{2A{ϵ}_0}{3d}=\frac{2}{3}C$

Q-5.Find out equivalent capacitance between A and B.

Solution:

Let charge distributions on plates as shown:

$C=\frac{Q}{V}=\frac{x+y}{V_{AB}}$

$\text{Potential of 1 and 4 is same:}$

$\frac{y}{A{ϵ}_0}=\frac{2x}{A{ϵ}_0}\Rightarrow y=2x$

$V=\left(\frac{2y+x}{A{ϵ}_0}\right)d$

$C=\frac{(x+2x)A{ϵ}_0}{(5x)d}=\frac{3A{ϵ}_0}{5d}$

Q-6.In a parallel plate capacitor setup, the lower plate is fixed on a rigid support, while the upper plate is suspended from one arm of a balance. Initially, the plates are connected with a thin wire and then disconnected after achieving the same potential. The balance is counterpoised at this stage. A voltage of V=5000V is then applied across the plates. The separation between the plates is d=5 mm , and the area of each plate is $A=100c{m}^2$ . Assuming all components except the plates are massless and non-conducting, calculate the additional mass that must be placed on the other side of the balance to restore equilibrium.

Solution: The electric force between the plates will be balanced by the additional weight

$mg=\frac{Q^2}{2A{ϵ}_0}=\frac{C^2{V}^2}{2A{ϵ}_0}$

$mg=\frac{{ϵ}_{o}A{V}^2}{2d^2}$

$m=\frac{{ϵ}_{o}A{V}^2}{2d^{2}g}=\frac{{ϵ}_o\times 100\times 10^{-4}\times (5000{)}^2}{2\times (5\times 10^{-4}{)}^3\times 10}$

$m=4.425g$

Q-7.One plate of a capacitor is unchanged and the other is connected to a spring as shown in Fig. The area of both the plates is A. In steady state (equilibrium), separation between the plates is 0.8 d (spring was unstretched and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately.

Solution:

$mg=\frac{Q^2}{2A{ϵ}_0}=K(0.2d)$

$\frac{(\frac{{ϵ}_{0}A}{0.8d}E{)}^2}{2A{ϵ}_0}=0.2Kd$

$K=\frac{3.9{ϵ}_{0}A{E}^2}{d^3}$

Q-8.Three dielectric slabs are inserted between the plates of a parallel plate capacitor of plate area A , as shown in the figure.

• Slab 1: Dielectric constant $K_1$​, area $A_1$, thickness d
• Slab 2: Dielectric constant $K_2$, area  $A_2$​, thickness $d_1$
• Slab 3: Dielectric constant $K_3$, area $A_2$​, thickness $d_2$

The total distance between the capacitor plates is $d=d_1+d_2$​. Find the equivalent capacitance between points A and B.

Solution:

It is equivalent to $C=C_1+\frac{C_2{C}_3}{C_2+C_3}$

$C=\frac{A{K}_1{\epsilon }_0}{d_1+d_2}$ + $\frac{\frac{A_2{K}_2{ϵ}_0}{d_1}.\frac{A_2{K}_3{ϵ}_0}{d_2}}{\frac{A_2{K}_2{ϵ}_0}{d_1}+\frac{A_2{K}_3{ϵ}_0}{d_2}}$

$C=\frac{A{K}_1{\epsilon }_0}{d_1+d_2}$+ $\frac{A_2^2{K}_2{K}_3{ϵ}_0^2}{A_2{K}_2{ϵ}_0{d}_2+A_2{K}_3{ϵ}_0{d}_1}$

$C=\frac{A{K}_1{\epsilon }_0}{d_1+d_2}+\frac{A{K}_2{K}_1{\epsilon }_0}{K_2{d}_2+K_1{d}_1}$

Q-9.Two dielectric slabs with dielectric constants $K_{1}and{K}_2$​​, and thicknesses $d_{1}and{d}_2$respectively, are inserted between the plates of a parallel plate capacitor. Each slab fully covers the plate area A, and the total separation between the plates is $d_1+d_2$ ​. Determine the equivalent capacitance between points A and B.

Solution:

$C=\frac{\sigma A}{V}(C=\frac{Q}{V};q=\sigma A)$

$V=E_1{d}_1+E_2{d}_2=\frac{\sigma d_1}{K_1{\epsilon }_0}+\frac{\sigma d_2}{K_2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}\left(\frac{d_1}{K_1}+\frac{d_2}{K_2}\right)$

$C=\frac{A{\epsilon }_0}{\frac{d_1}{K_1}+\frac{d_2}{K_2}}$

$\frac{1}{C}=\frac{d_1}{A{K}_1{\epsilon }_0}+\frac{d_2}{A{K}_2{\epsilon }_0}$

Q-10.If potential of A is 5V, then potential of B in volt is 

Solution:

$1(8-V_B)+1(8-V_B)+1(5-V_B)=0$

$21=3V_B$

$V_B=7V$

Q-11.An isolated parallel plate capacitor consists of two metallic plates, each of area A , separated by a distance d . A dielectric slab of width t and dielectric constant K is inserted between the plates such that its faces are parallel to the plates and its surface area matches that of the plates.

1. Determine the capacitance of the system with the dielectric slab partially filling the space.
2. If K=2, find the ratio $\frac{t}{d}$ for which the capacitance becomes $\frac{3}{2}$ times that of the same capacitor when it is completely filled with air.
3. Assuming the charge q on the plates remains constant, calculate the ratio of the electrostatic energy stored in the two cases (with and without the dielectric slab). Also, explain the physical reason for any change in energy.

Solution:

$C^{\prime }=\frac{A{\epsilon }_0}{d-t+\frac{t}{k}}$

• $\text{With dielectric, }{C}^{\prime }=\frac{A{\epsilon }_0}{d-t+\frac{t}{k}}=\frac{3C}{2}=\frac{3}{2}\frac{{\epsilon }_{0}A}{d}⟹\frac{d}{t}=\frac{3}{2}$
• $\text{Energy }=\frac{Q^2}{2C}$

            $\text{Energy in 1st case, }{E}_1=\frac{q^2}{2C}$

            $\text{Energy in 2nd case, }{E}_2=\frac{q^2}{2C^{\prime }}$

$\Rightarrow \frac{E_1}{E_2}=\frac{C^{\prime }}{C}=\frac{3}{2}$

• $\Delta E=E_2-E_1=\frac{q^2}{2C^{\prime }}-\frac{q^2}{2C}=\frac{q^2}{2}(\frac{2}{3C}-\frac{1}{C})=-\frac{q^2}{6C}=-\frac{q^{2}d}{6A{\epsilon }_0}$