| ⇦ |
| ⇨ |
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: - The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz.
- Two concentric spheres kept in air have radii R and r.
- A horizontal straight conductor kept in north-south direction falls under gravity. Then
- A condenser of 250 μF is connected in parallel to a coil of inductance of 0.16 mH
- The depletion layer in the p-n junction region is caused by
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The threshold frequency for a photosensitive metal is 3.3 x 10¹⁴ Hz.
- Two concentric spheres kept in air have radii R and r.
- A horizontal straight conductor kept in north-south direction falls under gravity. Then
- A condenser of 250 μF is connected in parallel to a coil of inductance of 0.16 mH
- The depletion layer in the p-n junction region is caused by
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