Question

4. What will be the amount of work done to rotate an electric dipole of moment 2.0 x 10-20 C.m through an angle of 180° in an electric field of intensity 1.2 x 104 v.m-l? [4.8 x 10-16]​

Answer 1

Answer:

P=3×10

−8

Cm;E=10

4

N/C

At stable equilibrium (θ

1

)=0

∘

At unstable equilibrium (θ

2

)=180

∘

Work done in a rotating dipole is given by:

W=PE(cosθ

1

−cosθ

2

)=(3×10

−8

)(10

4

)[cos0

∘

−cos180

∘

]=3×10

−4

[1−(−1)]

W=6×10

−4

J.

Answer 2

Answer:

so sorry but I can't follow anyone even I can't follow my bestie sorry

Explanation:

P=3×10

−8

Cm;E=10

4

N/C

At stable equilibrium (θ

1

)=0

∘

At unstable equilibrium (θ

2

)=180

∘

Work done in a rotating dipole is given by:

W=PE(cosθ

1

−cosθ

2

)=(3×10

−8

)(10

4

)[cos0

∘

−cos180

∘

]=3×10

−4

[1−(−1)]

W=6×10

−4

J

hope it's helpful

❤️sam❤️