back to directory
NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEEasy

Question

An electric dipole of moment pp is placed in an electric field of intensity EE. The dipole acquires a position such that the axis of the dipole makes an angle θ\theta with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ=90\theta=90^\circ, the torque and the potential energy of the dipole will respectively be:

A

pEsinθ,pEcosθpE\sin\theta, -pE\cos\theta

B

pEsinθ,2pEcosθpE\sin\theta, -2pE\cos\theta

C

pEsinθ,2pEcosθpE\sin\theta, 2pE\cos\theta

D

pEcosθ,pEsinθpE\cos\theta, -pE\sin\theta

Step-by-Step Solution

  1. Torque (τ\tau): When an electric dipole of moment p\mathbf{p} is placed in a uniform electric field E\mathbf{E}, it experiences a torque given by the vector product τ=p×E\tau = \mathbf{p} \times \mathbf{E}. The magnitude of this torque is τ=pEsinθ\tau = pE\sin\theta [NCERT Class 12, Physics Part 1, Sec 1.11, Eq. 1.22].
  2. Potential Energy (UU): The potential energy of the dipole is the work done in rotating the dipole from a reference position (where potential energy is zero) to the current position θ\theta. The standard reference position is taken at θ0=90\theta_0 = 90^\circ (perpendicular to the field). The work done is U=90θτdθ=90θpEsinθdθ=pE[cosθ]90θ=pEcosθU = \int_{90^\circ}^{\theta} \tau d\theta = \int_{90^\circ}^{\theta} pE\sin\theta d\theta = -pE [\cos\theta]_{90^\circ}^{\theta} = -pE\cos\theta [NCERT Class 12, Physics Part 1, Sec 2.8.3, Eq. 2.32].

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEelectricdipolemomentplacedelectric

More ELECTROSTATIC POTENTIAL AND CAPACITANCE Questions

View all

The variation of electrostatic potential with radial distance $r$ from the centre of a positively charged metallic thin shell of radius $R$ is given by the graph:

A.Option 1
B.Option 2
C.Option 3
D.Option 4
EasySolve

Six charges +q, -q, +q, -q, +q and -q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q₀ to the centre of the hexagon from infinity is: (ε₀ - permittivity of free space)

A.zero
B.\frac{-q^2}{4\pi\varepsilon_0 d}
C.\frac{-q^2}{4\pi\varepsilon_0 d}(3-\frac{1}{\sqrt{2}})
D.\frac{-q^2}{4\pi\varepsilon_0 d}(6-\frac{1}{\sqrt{2}})
EasySolve

A, B and C are three points in a uniform electric field. The electric potential is:

A.maximum at B
B.maximum at C
C.same at all the three points A, B and C
D.maximum at A
EasySolve

The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11}$ V. The electric field at that point is:

A.$4\pi\epsilon_0 Q \times 10^{22} \text{ V/m}$
B.$12\pi\epsilon_0 Q \times 10^{20} \text{ V/m}$
C.$4\pi\epsilon_0 Q \times 10^{20} \text{ V/m}$
D.$12\pi\epsilon_0 Q \times 10^{22} \text{ V/m}$
MediumSolve

Four point charges -Q, -q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero, is

A.Q = -q
B.Q = -1/q
C.Q = q
D.Q = 1/q
EasySolve

If potential in a region is expressed as $V(x,y,z) = 6xy - y + 2yz$, the electric field at point $(1, 1, 0)$ is:

A.$-(3\hat{i} + 5\hat{j} + 3\hat{k})$
B.$-(6\hat{i} + 5\hat{j} + 2\hat{k})$
C.$-(2\hat{i} + 3\hat{j} + \hat{k})$
D.$-(6\hat{i} + 9\hat{j} + \hat{k})$
MediumSolve

Two spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is:

A.a/b
B.b/a
C.a²/b²
D.b²/a²
MediumSolve

Two metal spheres, one of radius $R$ and the other of radius $2R$ respectively have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them?

A.$\sigma_1 = \frac{5}{6}\sigma, \sigma_2 = \frac{5}{6}\sigma$
B.$\sigma_1 = \frac{5}{2}\sigma, \sigma_2 = \frac{5}{6}\sigma$
C.$\sigma_1 = \frac{5}{2}\sigma, \sigma_2 = \frac{5}{3}\sigma$
D.$\sigma_1 = \frac{5}{3}\sigma, \sigma_2 = \frac{5}{6}\sigma$
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →