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
−cosθ
)=(3×10
)(10
)[cos0
−cos180
]=3×10
−4
[1−(−1)]
W=6×10
J.
so sorry but I can't follow anyone even I can't follow my bestie sorry
Explanation:
J
hope it's helpful
❤️sam❤️
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Answers & Comments
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:
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❤️