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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: - In Young’s double slit experiment, the fringe width is 1×10⁻⁴ m, if the distance
- The transverse nature of electromagnetic waves is proved by which of the following?
- A plane wave of wavelength 6250Å is incident normally on a slit of width 2×10⁻² cm.
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- A block of mass 0.50 kg is moving with a speed of 2.00 ms⁻¹ on a smooth surface.
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double slit experiment, the fringe width is 1×10⁻⁴ m, if the distance
- The transverse nature of electromagnetic waves is proved by which of the following?
- A plane wave of wavelength 6250Å is incident normally on a slit of width 2×10⁻² cm.
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- A block of mass 0.50 kg is moving with a speed of 2.00 ms⁻¹ on a smooth surface.
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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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