Find the potential energy of an electric dipole length 4 cm each experiences a torque of 4- 3 Nm placed in a from electric field angle 60- q - 1 - nc-

Given, #10; #10;Torque, τ =4√(3) N at 60^o #10; #10; τ =p⃗×E⃗=4√(3) #10; #10; pEsin60^o=4√(3) #10; #10; pE=8 Nm ....... ...

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Standard XII

Physics

Potential Energy - External Viewpoint

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Question

Find the potential energy of an electric dipole (length 4 cm) each experiences a torque of $4 \sqrt{3}$ Nm placed in a from electric field angle $60^{\circ} q = 1 + n c .$

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Solution

Given,

Torque, $τ = 4 \sqrt{3} N a t 60^{o}$

$τ = \to p \times \to E = 4 \sqrt{3}$

$p E sin 60^{o} = 4 \sqrt{3}$

$p E = 8 N m . . . . . . . (1)$

Potential Energy.

$U (θ) = p E (cos θ_{0} - cos θ_{1})$

$U (60^{o}) = 8 (cos 0^{o} - cos 60^{o}) = 4 J$

Hence, potential energy is $4 J$

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Find the potential energy of an electric dipole (length 4 cm) each experiences a torque of 4√3 Nm placed in a from electric field angle 60∘q=1+nc.

Given, Torque,τ=4√3Nat60o τ=→p×→E=4√3 pEsin60o=4√3 pE=8Nm.......(1) Potential Energy. U(θ)=pE(cosθ0−cosθ1) U(60o)=8(cos0o−cos60o)=4J Hence, potential energy is 4J

Find the potential energy of an electric dipole (length 4 cm) each experiences a torque of $4 \sqrt{3}$ Nm placed in a from electric field angle $60^{\circ} q = 1 + n c .$

Given,

Torque, $τ = 4 \sqrt{3} N a t 60^{o}$

$τ = \to p \times \to E = 4 \sqrt{3}$

$p E sin 60^{o} = 4 \sqrt{3}$

$p E = 8 N m . . . . . . . (1)$

Potential Energy.

$U (θ) = p E (cos θ_{0} - cos θ_{1})$

$U (60^{o}) = 8 (cos 0^{o} - cos 60^{o}) = 4 J$

Hence, potential energy is $4 J$