NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Ans: It is provided that,

First charge, ${{q}_{1}}=5\times {{10}^{-8}}C$

Second charge, \[{{q}_{2}}=-3\times {{10}^{-8}}C\]

Distance between the two given charges, $d=16cm=0.16m$

Case 1. When point P is inside the system of two charges.

Consider a point named P on the line connecting the two charges.

r is the distance of point P from ${{q}_{1}}$.

Potential at point P will be,

$V=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}r}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(d-r)}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

But $V=0$so,

$0=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}r}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(d-r)}$

$\Rightarrow \frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}r}=-\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(d-r)}$

$\Rightarrow \frac{{{q}_{1}}}{r}=-\frac{{{q}_{2}}}{(d-r)}$

\[\Rightarrow \frac{5\times {{10}^{-8}}}{r}=-\frac{-3\times {{10}^{-8}}}{(0.16-r)}\]

\[\Rightarrow \frac{0.16}{r}=\frac{8}{5}\]

We get,

$r=0.1m=10cm$

Therefore, the potential is zero at \[10\text{ }cm\] distance from the positive charge.

Case 2. When point P is outside the system of two charges.

Potential at point P will be,

$V=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}s}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(s-d)}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

But $V=0$so,

$0=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}s}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(s-d)}$

$\Rightarrow \frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}s}=-\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(s-d)}$

$\Rightarrow \frac{{{q}_{1}}}{s}=-\frac{{{q}_{2}}}{(s-d)}$

\[\Rightarrow \frac{5\times {{10}^{-8}}}{s}=-\frac{-3\times {{10}^{-8}}}{(s-0.16)}\]

\[\Rightarrow \frac{0.16}{s}=\frac{2}{5}\]

We get,

$s=0.4m=40cm$

Therefore, the potential is zero at \[40\text{ }cm\] distance from the positive charge.

2. A regular hexagon of side \[10\text{ }cm\] has a charge \[5\mu C\] at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans: The given figure represents six equal charges, $q=5\times {{10}^{-6}}C$, at the hexagon's vertices.

Sides of the hexagon, $AB=BC=CD=DE=EF=FA=10cm$

The distance of O from each vertex, $d=10cm$

Electric potential at point O,

$V=\frac{6q}{4\pi {{\varepsilon }_{o}}d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow V=\frac{6\times 9\times {{10}^{9}}\times 5\times {{10}^{-6}}}{0.1}$

$\Rightarrow V=2.7\times {{10}^{6}}V$

Clearly, the potential at the hexagon’s centre is $2.7\times {{10}^{6}}V$.

3. Two charges \[2\mu C\] and \[-2\mu C\] are placed at points A and B, \[\mathbf{6}\text{ }\mathbf{cm}\] apart. 

a) Identify an equipotential surface of the system. 

Ans: The given figure represents two charges.

An equipotential surface is defined as that plane on which electric potential is equal at every point. One such plane is normal to line AB. The plane is placed at the mid-point of line AB because the magnitude of charges is equal.

b) What is the direction of the electric field at every point on this surface?

Ans: The electric field's direction is perpendicular to the plane in the line AB direction at every location on this surface.

4. A spherical conductor of radius \[\mathbf{12}\text{ }\mathbf{cm}\] has a charge of $1.6\times {{10}^{-7}}C$ distributed uniformly on its surface. What is the electric field, 

a) inside the sphere?

Ans: It is provided that,

Spherical conductor’s radius, \[r=12cm=0.12m\]

The charge is evenly distributed across the conductor. The electric field within a spherical conductor is zero because the total net charge within a conductor is zero.

b) just outside the sphere?

Ans: Just outside the conductor, Electric field E is given by

$E=\frac{q}{4\pi {{\varepsilon }_{o}}{{r}^{2}}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow E=\frac{1.6\times {{10}^{-7}}\times 9\times {{10}^{9}}}{{{(0.12)}^{2}}}$

$\Rightarrow E={{10}^{5}}N{{C}^{-1}}$

Clearly, the electric field just outside the sphere is ${{10}^{5}}N{{C}^{-1}}$.

c) at a point \[\mathbf{18}\text{ }\mathbf{cm}\] from the centre of the sphere?

Ans: Let electric field at a given point which is \[18\text{ c}m\] from the sphere centre = ${{E}_{1}}$

Distance of the given point from the centre, $d=\mathbf{18}\text{ }\mathbf{cm}=0.18m$

The formula for electric field is given by,

${{E}_{1}}=\frac{q}{4\pi {{\varepsilon }_{o}}{{d}^{2}}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow E=\frac{1.6\times {{10}^{-7}}\times 9\times {{10}^{9}}}{{{(0.18)}^{2}}}$

$\Rightarrow E=4.4\times {{10}^{4}}N{{C}^{-1}}$

Therefore, the electric field at a given point \[18\text{ }cm\] from the sphere centre is $4.4\times {{10}^{4}}N{{C}^{-1}}$.

5. A parallel plate capacitor with air between the plates has a capacitance of \[~8pF\]. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant \[6\]?

Ans: It is provided that,

Capacitance between the capacitor’s parallel plates, \[C=8pF\]

Originally, the distance separating the parallel plates was d, and the air was filled in it.

Dielectric constant of air, \[k=1\]

The formula for Capacitance is given by:

$C=\frac{k{{\varepsilon }_{o}}A}{d}$

Here, \[k=1\], so,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

A is the area of each plate.

If the distance separating the plates is decreased to half and the substance has a dielectric constant of \[6\] filled in between the plates.

Then,

$k'=6,d'=\frac{d}{2}$

Hence, capacitor’s capacitance becomes,

$C'=\frac{k'{{\varepsilon }_{o}}A}{d'}$

$\Rightarrow C'=\frac{6{{\varepsilon }_{o}}A}{\frac{d}{2}}$

$\Rightarrow C'=12C$

$\Rightarrow C'=12\times 8=96pF$

Therefore, the capacitance when the substance of dielectric constant $6$ is filled between the plates is \[96\text{ }pF\].

6. Three capacitors each of capacitance \[\mathbf{9}\text{ }\mathbf{pF}\] are connected in series. 

a) What is the total capacitance of the combination? 

Ans: It is provided that,

Capacitance of each three capacitors, \[C=9pF\]

The formula for equivalent capacitance $(~C')$ of the capacitors’ series combination is given by

$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$

$\Rightarrow \frac{1}{C'}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$

$\Rightarrow \frac{1}{C'}=\frac{3}{9}$

$\Rightarrow C'=3pF$

Clearly, total capacitance of the combination of the capacitors is \[3pF\].

b) What is the potential difference across each capacitor if the combination is connected to a \[\mathbf{120}\text{ }\mathbf{V}\] supply?

Ans: Provided that,

Supply voltage, \[V=120\text{ }V\]

Potential difference \[(V')\] across each capacitor will be one-third of the supply voltage.

$V'={}^{120}/{}_{3}=40V$

Clearly, the potential difference across each capacitor is \[40\text{ }V\].

7. Three capacitors of capacitance \[2pF,\text{ }3\text{ }pF\text{ }and\text{ }4pF\] are connected in parallel. 

a) What is the total capacitance of the combination?

Ans: Provided that,

Capacitances of the given capacitors are, 

\[{{C}_{1}}=2pF\text{ };\text{ }{{C}_{2}}=3pF\text{ };\text{ }{{C}_{3}}=4pF\]

The formula for equivalent capacitance $(~C')$ of the capacitors’ parallel combination is given by

\[C'={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]

\[\Rightarrow C'=2+3+4=9pF\]

Therefore, total capacitance of the combination is \[9pF.\]

b) Determine the charge on each capacitor if the combination is connected to a \[100\text{ }V\] supply.

Ans: We have,

Supply voltage, \[V=100\text{ }V\]

Charge on a capacitor with capacitance C and potential difference V is given by,

$q=CV$……(i)

For $C=2pF$,

Charge $=VC=100\times 2=200pF$

For $C=3pF$,

Charge $=VC=100\times 3=300pF$

For $C=4pF$,

Charge $=VC=100\times 4=400pF$

8. In a parallel plate capacitor with air between the plates, each plate has an area of $6\times {{10}^{-3}}{{m}^{2}}$ and the distance between the plates is \[3\text{ }mm\]. Calculate the capacitance of the capacitor. If this capacitor is connected to a \[100\text{ }V\] supply, what is the charge on each plate of the capacitor?

Ans: It is provided that,

Area of parallel plate capacitor’s each plate, $A=6\times {{10}^{-3}}{{m}^{2}}$

Distance separating the plates, \[d=3mm=3\times {{10}^{-3}}m\]

Supply voltage, \[V=100\text{ }V\] 

The formula for parallel plate capacitor’s Capacitance is given by,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$\Rightarrow C=\frac{8.854\times {{10}^{-12}}\times 6\times {{10}^{-3}}}{3\times {{10}^{-3}}}$

\[\Rightarrow C=17.71\times {{10}^{-12}}F\]

\[\Rightarrow C=17.71pF\]

The formula for Potential V is related with charge q and capacitance C is given by,

$V=\frac{q}{C}$

$\Rightarrow q=CV=100\times 17.71\times {{10}^{-12}}$

$\Rightarrow q=1.771\times {{10}^{-9}}C$

Clearly, the capacitor's capacitance is \[17.71\text{ }pF\] and charge on each plate is $1.771\times {{10}^{-9}}C$.

9. Explain what would happen if in the capacitor given in Exercise 8, a \[3\text{ }mm\] thick mica sheet (of dielectric constant \[=6\]) were inserted between the plates, 

a) while the voltage supply remained connected.

Ans: It is provided that,

Mica sheet’s Dielectric constant, k = 6 

Initial capacitance, \[C=17.71\times {{10}^{-12}}F\]

New capacitance, \[C'=kC=6\times 17.71\times {{10}^{-12}}=106pF\]

Supply voltage, \[V=100V\]

New charge, $q'=C'V'=106\times 100pC=1.06\times {{10}^{-8}}C$

Potential across the plates will remain \[100\text{ }V\].

c) after the supply was disconnected.

Ans: It is provided that,

Mica sheet’s Dielectric constant, k = 6 

Initial capacitance, \[C=17.71\times {{10}^{-12}}F\]

New capacitance, \[C'=kC=6\times 17.71\times {{10}^{-12}}=106pF\]

If supply voltage is disconnected, then there will be no influence on the charge amount on the plates. 

The formula for potential across the plates is given by,

$V'=\frac{q}{C'}$

$V'=\frac{1.771\times {{10}^{-9}}}{106\times {{10}^{-12}}}=16.7V$

The potential across the plates when the supply was removed is $16.7V$.

10. A \[12pF\] capacitor is connected to a \[50\text{ }V\] battery. How much electrostatic energy is stored in the capacitor?

Ans: It is provided that,

Capacitance of the capacitor, \[C=12\times {{10}^{-12}}F\]

Potential difference, \[V=50\text{ }V\]

The formula for stored electrostatic energy in the capacitor is given by,

$E=\frac{1}{2}C{{V}^{2}}$

$\Rightarrow E=\frac{1}{2}\times 12\times {{10}^{-12}}\times {{50}^{2}}$

$\Rightarrow E=1.5\times {{10}^{-8}}J$

Therefore, the stored electrostatic energy in the capacitor is $1.5\times {{10}^{-8}}J$.

11. A \[600\text{ }pF\] capacitor is charged by a \[200\text{ }V\] supply. It is then disconnected from the supply and is connected to another uncharged \[600\text{ }pF\] capacitor. How much electrostatic energy is lost in the process?

Ans: It is provided that,

Capacitance of the capacitor, \[C=600\text{ }pF\]

Potential difference, \[V=200\text{ }V\]

The formula for stored electrostatic energy in the capacitor is given by,

$E=\frac{1}{2}C{{V}^{2}}$

$\Rightarrow E=\frac{1}{2}\times 600\times {{10}^{-12}}\times {{200}^{2}}$

$\Rightarrow E=1.2\times {{10}^{-5}}J$

If supply is removed from the capacitor and another capacitor of capacitance \[C=600\text{ }pF\] is joined to it, then equivalent capacitance $(C')$ of the series combination is given by

$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}$

$\Rightarrow \frac{1}{C'}=\frac{1}{600}+\frac{1}{600}$

$\Rightarrow \frac{1}{C'}=\frac{2}{600}$

$\Rightarrow C'=300pF$

New electrostatic energy will be,

$E'=\frac{1}{2}C'{{V}^{2}}$

$\Rightarrow E'=\frac{1}{2}\times 300\times {{10}^{-12}}\times {{200}^{2}}$

$\Rightarrow E'=0.6\times {{10}^{-5}}J$

Loss in electrostatic energy $=E-E'$

$\Rightarrow E-E'=1.2\times {{10}^{-5}}-0.6\times {{10}^{-5}}=0.6\times {{10}^{-5}}J$

$\Rightarrow E-E'=6\times {{10}^{-6}}J$

Clearly, the lost electrostatic energy in the process is $6\times {{10}^{-6}}J$.

Electrostatic Potential and Capacitance Chapter Summary - Class 12 NCERT Solutions

• Electric Potential: Electric potential at a point in an electric field is the amount of work done in bringing a unit positive charge from infinity to the point.   

(i) Electric potential is a scalar quantity.

(ii) S.I. unit: Volt(V).  

(iii) A positive charge in a field moves from higher potential to lower potential whereas an electron moves from lower potential to higher potential when left free.   

(iv) Work done in moving a charge q through a potential difference V is W = qV joule.

• If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given by

$V(r)-\frac{1}{4\pi\varepsilon _0}\frac{Q}{r}$

For a charge configuration q1, q2, ..., qn with position vectors r1 , r2 , ... rn, the potential at a point P is given by the superposition principle

$V(r)-\frac{1}{4\pi\varepsilon _0}\left ( \frac{q_1}{r_{1P}}\frac{q_2}{r_{2P}}+...+\frac{q_n}{r_{nP}} \right )$

Where r1P is the distance between q1 and P, and so on.

• Relation Between Electric Field and Potential:

$V=-\iint_E^{\amalg} \cdot d \stackrel{凶}{\mathrm{凶}}$

$\Rightarrow \mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dr}}$

• An equipotential surface is a surface over which potential has a constant value. For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces. The electric field E at a point is perpendicular to the equipotential surface through the point. E is in the direction of the steepest decrease of potential.

• Electric potential at any point due to an electric dipole $V \equiv \frac{k p \cos \theta}{r^2}$; $\theta$ ; is the angle

• Between $\stackrel{\mathrm{w}}{\mathrm{p}} and\, \stackrel{\mathrm{w}}{\mathrm{w}}$

• Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. Potential energy of two charges q1, q2 at r1, r2 is given by $U=\frac{1}{4\pi\varepsilon _0}\frac{q_1q_2}{r_{12}}$ where r12 is distance between q1 and q2. The potential energy of a charge q in an external potential V(r) is qV(r). The potential energy of a dipole moment p in a uniform electric field E is –p.E.

• Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by $E=\frac{\sigma}{\varepsilon _0}n$ where n is the unit vector along the outward normal to the surface and σ is the surface charge density. Charges in a conductor can reside only at its surface. Potential is constant within and on the surface of a conductor. In a cavity within a conductor (with no charges), the electric field is zero.

• A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. C is determined purely geometrically, by the shapes, sizes and relative positions of the two conductors. The unit of capacitance is farad:, 1 F = 1 C V–1. For a parallel plate capacitor (with vacuum between the plates), $C=\varepsilon _0\frac{A}{d}$ where A is the area of each plate and d the separation between them.

• If the medium between the plates of a capacitor is filled with an insulating substance (dielectric), the electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarisation, gives rise to a field in the opposite direction. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. Consequently, the capacitance C increases from its value C0 when there is no medium (vacuum), 

C = KC0

where K is the dielectric constant of the insulating substance.

• For capacitors in the series combination, the total capacitance C is given by $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...$ In the parallel combination, the total capacitance C is: 

$C=C_1+C_2+C_3+...$

where C1, C2, C3 ... are individual capacitances.

• The energy U stored in a capacitor of capacitance C, with charge Q and voltage V is

$U=\frac{1}{2}QV=\frac{1}{2}CV^2=\frac{1}{2}\frac{Q^2}{C}$ 

The electric energy density (energy per unit volume) in a region with electric field is (1/2)ε0E2.

Important Formulas Covered in the NCERT Solutions for Class 12 Physics Chapter 2

Students who are preparing for the Physics 12th board exams can go through the below given important formulas for brushing up their concepts.

1. Electric Potential Energy ΔU = W  

Here, ΔU represents changes in potential energy and W represents the work done by electric lines of force.

2. Electric dipole P = qd

Here, q is the test charge and d represents the vector joining two charge points.

3. Capacitance of the capacitor C = \[\frac{q}{v}\]

The unit of capacitance is farad.

4. Energy stored in the capacitor E = \[\frac{QV}{2}\] or \[\frac{CV^{2}}{2}\] 

Students can download the free PDF of formulas used in the NCERT Class 12 Physics Chapter 2 from Vedantu for last-minute revision.

Overview of Deleted Syllabus for CBSE Class 12 Physics Electrostatic Potential and Capacitance

Conclusion

NCERT Class 12 Maths Chapter 2 Solutions on Electrostatic Potential and Capacitance provided by Vedantu provides a comprehensive understanding of electric potential energy, potential difference, and the functioning of capacitors. This knowledge is crucial for both academic success and practical applications in technology and engineering.

The solution cover exercise in the NCERT textbook offers clear explanations and step-by-step methods for solving problems. Important points to focus on include understanding the behaviour of electric fields, potential, and energy storage in capacitors. From previous year's question papers, typically around 7 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills.

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Chapter-Specific NCERT Solutions for Class 12 Physics

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