An electric dipole moment p→=(2.0i^+3.0j^) μC . It is placed in a uniform electric field E→=(3.0i^+2.0k^)×105NC−1

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The torque that E→ exerts on p→ is (0.6i^−0.4j^−0.9k^) Nm

The potential energy of the dipole is – 0.6J

The potential energy of the dipole is 0.6J

If the dipole is rotated in the electric field, the maximum potential energy of the dipole is 1.3J

answer is Ü.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

P=(2i+3j)10−6 and E=(3i+2k)105 τ¯=P¯×E¯=(0.6i−0.4j−0.9k)U=−P.E=2×3×10−1=−0.6Umax=|P||E|=13×13×10−1=1.3J