Answer the following. (i) Draw equipotential surfaces for an electric dipole. (ii) Two point charges q_1 and q_2 are located at r_1 and r_2 respectively in an external electric field E. Obtain an expression for the potential energy of the system. (iii) The dipole moment of a molecule is 10^-30 Cm. It is placed in an electric field E of 10^5 V/m such that its axis is along the electric field. The direction of E is suddenly changed by 60° at an instant. Find the change in the potential energy of the dipole, at that instant.

(ii) U = q_1V( r_1)+q_2V( r_2)+q_1q_24πε_0 r12; (iii) ΔU = 5.0 × 10⁻²⁶ J

PhysicsCBSE Class 12 · Board exam 2024, set 55/5/1 · 5 marks · Case-based

CBSE Class 12 Physics 2024: question 31(a)

Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, section E, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Case-basedCompetency-basedElectrostatic Potential and Capacitance

Answer the following.

(i)
Draw equipotential surfaces for an electric dipole.
0 marks
(ii)
Two point charges $q_{1}$ and $q_{2}$ are located at $r_{1}$ and $r_{2}$ respectively in an external electric field $E$ . Obtain an expression for the potential energy of the system.
0 marks
(iii)
The dipole moment of a molecule is $1 0^{- 30}$ Cm. It is placed in an electric field $E$ of $1 0^{5}$ V/m such that its axis is along the electric field. The direction of $E$ is suddenly changed by $6 0^{\circ}$ at an instant. Find the change in the potential energy of the dipole, at that instant.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Equipotential surfaces are always perpendicular to field lines.

Hint 2

Add the work done to bring each charge against the external field, and the work done against their mutual interaction.

Hint 3

Use U = −pE cos θ at the initial and final angles.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Equipotential surfaces of a dipole (figure in words):
- Draw $+ q$ and $- q$ a short distance apart.
- Around each charge draw closed, distorted oval surfaces that are crowded near the charge.
- The surfaces around $+ q$ are at positive potential and those around $- q$ at negative potential.
- The perpendicular bisector plane of the dipole is the zero-potential surface.
- Electric field lines, from $+ q$ to $- q$ , cut every surface at right angles.
1 mark
(ii) Bring $q_{1}$ from infinity to $r_{1}$ in the external field. Work done:
$W_{1} = q_{1} V (r_{1})$ ...(1)
½ mark
Bring $q_{2}$ from infinity to $r_{2}$ . Work done against the external field:
$W_{2} = q_{2} V (r_{2})$ ...(2)
½ mark
Work done on $q_{2}$ against the field of $q_{1}$ :
$W_{12} = \frac{q _{1} q _{2}}{4 π ε _{0} r _{12}}$ ...(3), where $r_{12} = ∣ r_{1} - r_{2} ∣$
½ mark
Potential energy of the system = total work done:
$U = q_{1} V (r_{1}) + q_{2} V (r_{2}) + \frac{q _{1} q _{2}}{4 π ε _{0} r _{12}}$
½ mark
(iii) Potential energy of a dipole: $U = - pE cos θ$ .
The change as $θ$ goes from $0^{\circ}$ to $6 0^{\circ}$ is
$Δ U = pE (cos θ_{0} - cos θ_{1}) = pE (cos 0^{\circ} - cos 6 0^{\circ})$
1 mark
Substituting: $Δ U = 1 0^{- 30} \times 1 0^{5} \times (1 - 0.5)$
½ mark
$Δ U = 5.0 \times 1 0^{- 26}$ J (an increase in potential energy)
½ mark

Answer: (ii) $U = q_{1} V (r_{1}) + q_{2} V (r_{2}) + \frac{q _{1} q _{2}}{4 π ε _{0} r _{12}}$ ; (iii) ΔU = 5.0 × 10⁻²⁶ J

Where the marks are

Equipotential surfaces of a dipole drawn correctly (1 mark)
Work q₁V(r₁) (½ mark)
Work q₂V(r₂) (½ mark)
Mutual interaction term q₁q₂/4πε₀r₁₂ (½ mark)
Total potential energy expression (½ mark)
ΔU = pE(cos θ₀ − cos θ₁) (1 mark)
Substitution (½ mark)
Result 5 × 10⁻²⁶ J (½ mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)
Topic: Electric Potential and Potential Energy

Practise this chapterBoard-style questions from Electrostatic Potential and Capacitance, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2024: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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