| ⇦ |
| ⇨ |
An electric dipole of length 1 cm is placed with the axis making an angle of 30⁰ to an electric field of strength 10⁴ NC⁻¹. If it experiences a torque of Nm, the potential energy of the dipole is
Options
(a) 2.45 J
(b) 0.0245 J
(c) 245.0 J
(d) 24.5 J
Correct Answer:
24.5 J
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Two small spherical shells A and B are given positive charges of 9C and 4C
- In a Rutherford scattering experiment when a projectile of charge Z₁ and mass M₁
- The number of beta particles emitted by a radioactive substance is twice the number
- If t₁ ̷ ₂ is the half-life of a substance, then t₃ ̷ ₄ is the time in which
- A body with mass 5 kg is acted upon a force F=(-3i+4j) N.If its initial velocity at
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two small spherical shells A and B are given positive charges of 9C and 4C
- In a Rutherford scattering experiment when a projectile of charge Z₁ and mass M₁
- The number of beta particles emitted by a radioactive substance is twice the number
- If t₁ ̷ ₂ is the half-life of a substance, then t₃ ̷ ₄ is the time in which
- A body with mass 5 kg is acted upon a force F=(-3i+4j) N.If its initial velocity at
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Torque=PE sin¢
10√2= P×10^4 sin 30°
P=2√2×10^-3
U=PE cos 30°
U=2√2×10^-3× 10^4 cos 30°
=24.5 J
Torque=PE sin¢
10√2= P×10^4 sin 30°
P=2√2×10^-3
U=PE cos 30°
U=2√2×10^-3× 10^4 cos 30°
=24.5 J
Torque=PE sin¢
10√2= P×10^4 sin 30°
P=2√2×10^-3
U=PE cos 30°
U=2√2×10^-3× 10^4 cos 30°
=24.5 J