12th Physics Chapter 1 – 5 Mark Questions (Important & Repeated – TN Board)

Prepare for exams with 12th Physics Chapter 1 (Electrostatics) important 5 mark questions. This collection includes most repeated and previously asked questions based on TN Board exams, helping students revise key concepts quickly and score better.

Electrostatics – 5 Mark Questions

12th Physics Chapter 1 Important 5 Marks with Video Tutorial – TN Board Padidaa

Most Repeated 5 Marks

1.Electric field due to a dipole along Axial line ?(6)

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Consider an electric dipole placed on the x-axis, A point C is located at a distance of r from the midpoint O of the dipole on the axial line

\textbf{The electric field at a point C due to +q is}

\vec{E}_{+} = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r - a)^2} \hat{p}

\textbf{The electric field at a point C due to -q is}

\vec{E}_{-} = - \frac{1}{4\pi \varepsilon_0} \frac{q}{(r + a)^2} \hat{p}

\textbf{The total electric field}

\vec{E}_{tot} = \vec{E}_{+} + \vec{E}_{-}

\vec{E}_{tot} = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r - a)^2} \hat{p} - \frac{1}{4\pi \varepsilon_0} \frac{q}{(r + a)^2} \hat{p}

\vec{E}_{tot} = \frac{q}{4\pi \varepsilon_0} \left\{ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right\} \hat{p}

\vec{E}_{tot} = \frac{q}{4\pi \varepsilon_0} \left\{ \frac{4ra}{(r^2 - a^2)^2} \right\} \hat{p}

If the point C is very far away from the dipole (r >> a)}

\vec{E}_{tot} = \frac{1}{4\pi \varepsilon_0} \left\{ \frac{4aq}{r^3} \right\} \hat{p}

\textbf{since } 2aq \hat{p} = \vec{p}

\vec{E}_{tot} = \frac{1}{4\pi \varepsilon_0} \frac{2\vec{p}}{r^3}

It is important to note that the energy density depends only on the electric field and not on the size of the plates of the If the point C is chosen on the left side of the dipole, the total electric field is still in the direction of p vetor.

2.Electric field due to a dipole at a point on the Equatorial plane ?(2)

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Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane

\vec{E}_{+} \text{ and } \vec{E}_{-} \text{ can be resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.}

Electric field due to a dipole at a point on the Equatorial plane
Electric field due to a dipole at a point on the Equatorial plane

\vec{E}_{tot} = - \left| \vec{E}_{+} \right| \cos\theta \, \hat{p} - \left| \vec{E}_{-} \right| \cos\theta \, \hat{p}

\left| \vec{E}_{+} \right| = \left| \vec{E}_{-} \right| = \frac{1}{4\pi \varepsilon_0} \frac{q}{(r^2 + a^2)} \, \hat{p}

\vec{E}_{tot} = - \frac{1}{4\pi \varepsilon_0} \frac{q}{(r^2 + a^2)} \cos\theta \, \hat{p} - \frac{1}{4\pi \varepsilon_0} \frac{q}{(r^2 + a^2)} \cos\theta \, \hat{p}

= - \frac{1}{4\pi \varepsilon_0} \frac{2q \cos\theta}{(r^2 + a^2)} \, \hat{p}

\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}

= - \frac{1}{4\pi \varepsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} \, \hat{p}

= - \frac{1}{4\pi \varepsilon_0} \frac{\vec{p}}{(r^2 + a^2)^{3/2}}

\vec{p} = 2qa \, \hat{p}

\text{At very large distances } (r >> a)

= - \frac{1}{4\pi \varepsilon_0} \frac{\vec{p}}{r^3}

3.Explain the working of Van de Graaff Generator (5)

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  • Robert Van de Graaff designed a machine which produces a large amount of electrostatic potential difference, up to several million volts (107 V).
  • This Van de Graff generator works on the principle of electrostatic induction and action at points.
Van de Graaff Generator

Construction :

  • A large hollow spherical conductor is fixed on the insulating stand.
  • A pulley B is mounted at the centre of the hollow sphere and another pulley C is fixed at the bottom.
  • A belt made up of insulating materials like silk or rubber runs over both pulleys.
  • The pulley C is driven continuously by the electric motor.
  • Two comb shaped metallic conductors E and D are fixed near the pulleys.
  • The comb D is maintained at a positive potential of 104 V by a power supply.
  • The upper comb E is connected to the inner side of the hollow metal sphere.

Working :

  • Due to the high electric field near comb D, air between the belt and comb D gets ionized by the action of points.
  • The positive charges are pushed towards the belt and negative charges are attracted towards the comb D.
  • The positive charges stick to the belt and move up. When the positive charges on the belt reach the point near the comb E, the comb E acquires negative charge and the sphere acquires positive charge due to electrostatic induction.
  • As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere.
  • At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air.

The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

4.Electrostatic Potential at a point due to electric dipole (4) (Public Exam 2026)

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  • Consider two equal and opposite charges separated by a small distance 2a.
  • The point P is located at a distance r from the midpoint of the dipole.
  • Let θ be the angle between the line OP and dipole axis AB
  • Let r1 be the distance of point P from +q and r2 be the distance of point P from –q.
Electrostatic Potential at a point due to electric dipole

\text{Potential at P due to charge +q} = \frac{1}{4\pi \varepsilon_0} \frac{q}{r_1}

\text{Potential at P due to charge -q} = - \frac{1}{4\pi \varepsilon_0} \frac{q}{r_2}

\text{Total potential} = \frac{1}{4\pi \varepsilon_0} q \left\{ \frac{1}{r_1} - \frac{1}{r_2} \right\}

\text{By the cosine law for triangle BOP}

r_1^2 = r^2 + a^2 - 2ra \cos\theta

r_1^2 = r^2 \left[ 1 + \frac{a^2}{r^2} - \frac{2a}{r} \cos\theta \right]

\text{since } (r >> a), \frac{a^2}{r^2} \text{ will be very small and can be neglected}

\therefore r_1^2 = r^2 \left[ 1 - \frac{2a}{r} \cos\theta \right]

r_1 = r \left[ 1 - \frac{2a}{r} \cos\theta \right]^{1/2}

\frac{1}{r_1} = \frac{1}{r} \left[ 1 - \frac{2a}{r} \cos\theta \right]^{-1/2}

\text{Using Binomial Expansion}

\frac{1}{r_1} = \frac{1}{r} \left[ 1 - \left( -\frac{1}{2} \right) \frac{2a}{r} \cos\theta \right]

\frac{1}{r_1} = \frac{1}{r} \left[ 1 + \frac{a}{r} \cos\theta \right]

\text{Similarly applying the cosine law for triangle AOP}

\frac{1}{r_2} = \frac{1}{r} \left[ 1 - \frac{a}{r} \cos\theta \right]

= \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{r} \left( 1 + \frac{a}{r} \cos\theta \right) - \frac{1}{r} \left( 1 - \frac{a}{r} \cos\theta \right) \right]

= \frac{q}{4\pi \varepsilon_0} \frac{1}{r} \left[ 1 + \frac{a}{r} \cos\theta - 1 + \frac{a}{r} \cos\theta \right]

= \frac{q}{4\pi \varepsilon_0} \frac{1}{r} \left[ \frac{2a \cos\theta}{r} \right]

= \frac{1}{4\pi \varepsilon_0} \frac{2aq \cos\theta}{r^2}

= \frac{1}{4\pi \varepsilon_0} \frac{p \cos\theta}{r^2}

V = \frac{1}{4\pi \varepsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}

\textbf{Special cases}

\textbf{Case (i) If the point P lies on the axial line of the dipole on the side of +q, then } \theta = 0. \text{ Then the electric potential becomes}

V = \frac{1}{4\pi \varepsilon_0} \frac{p}{r^2} \quad [\cos\theta = 1]

\textbf{Case (ii) If the point P lies on the axial line of the dipole on the side of -q, then } \theta = 180^\circ \text{ Then}

V = - \frac{1}{4\pi \varepsilon_0} \frac{p}{r^2} \quad [\cos 180^\circ = -1]

\textbf{Case (iii) If the point P lies on the equatorial line of the dipole, then } \theta = 90^\circ. \text{ Hence}

V = 0

4.Electric field due to an infinitely long charged wire (Applications of Gauss Law) (Public Exam 2025)

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  • Consider an infinitely long straight wire having uniform linear charge density λ (charge per unit length).
  • Let P be a point located at a perpendicular distance r from the wire.
  • The electric field at the point P can be found using Gauss law.
Electric field due to an infinitely long charged wire

\textbf{Electric flux}

\phi_E = \oint \vec{E} \cdot d\vec{A}

\textbf{Total electric flux}

= \int \vec{E} \cdot d\vec{A} + \int \vec{E} \cdot d\vec{A} + \int \vec{E} \cdot d\vec{A}

\text{curved surface} \quad \text{top surface} \quad \text{bottom surface}

= \int E \, dA \cos 0^\circ + \int E \, dA \cos 90^\circ + \int E \, dA \cos 90^\circ

= \int E \, dA + 0 + 0

\text{Note: } \cos 0^\circ = 1, \quad \cos 90^\circ = 0

\phi_E = \int E \, dA = \frac{Q_{enc}}{\varepsilon_0}

Q_{enc} = \lambda L \quad [\text{charge density × length}]

E \int dA = \frac{\lambda L}{\varepsilon_0}

E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \quad [dA = \text{curved surface area of cylinder}]

E = \frac{1}{2\pi r L} \cdot \frac{\lambda L}{\varepsilon_0}

= \frac{1}{2\pi \varepsilon_0} \cdot \frac{\lambda}{r}

\text{in vector form}

\vec{E} = \frac{1}{2\pi \varepsilon_0} \cdot \frac{\lambda}{r} \, \hat{r}

\text{The electric field due to the infinite charged wire depends on } \frac{1}{r}

5.Electric field due to charged infinite plane sheet ( Applications of Gauss Law )

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  • Consider an infinite plane sheet of charges with uniform surface charge density σ.
  • Let P be a point at a distance of r from the sheet
Electric field due to charged infinite plane sheet

\text{Curved surface area, } \vec{E} \perp \vec{N} \text{ or } d\vec{A} \; \therefore \; E \, dA \cos 90^\circ = 0

\text{At P and P', } \vec{E} \parallel \vec{A} \text{ or } \vec{N} \; \therefore \; E \, dA \cos 0^\circ = E \, dA

\phi_E = \int_{P} E \, dA + \int_{P'} E \, dA = \frac{Q_{enc}}{\varepsilon_0}

Q_{enc} = \sigma A \quad [\text{charge density × area}]

2E \int dA = \frac{\sigma A}{\varepsilon_0}

2E A = \frac{\sigma A}{\varepsilon_0}

E = \frac{1}{2A} \cdot \frac{\sigma A}{\varepsilon_0}

E = \frac{\sigma}{2\varepsilon_0}

\text{in vector form}

\vec{E} = \frac{\sigma}{2\varepsilon_0} \, \hat{n}

electric field due to an infinite plane sheet of charge depends on the surface charge density and is independent of the distance r.

  • if σ > 0 the electric field at any point P is along outward perpendicular n
  • if σ < 0, the electric field points inward perpendicularly to the plane (-n ).

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