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Electric Charges and Fields - Dipole in a Uniform External Field

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An electric dipole consists of two equal and opposite charges +q+q and −q-q separated by a small distance 2a2a. The dipole moment is p⃗=q(2a)\vec{p} = q(2a) directed from negative to positive charge.

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When a dipole is placed in a uniform electric field E⃗\vec{E}, it experiences a net force F⃗net=qE⃗+(−qE⃗)=0\vec{F}_{net} = q\vec{E} + (-q\vec{E}) = 0. Thus, there is no translational motion.

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The two forces qE⃗q\vec{E} and −qE⃗-q\vec{E} form a couple, exerting a torque τ⃗\vec{\tau} that tends to align the dipole with the direction of the field.

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The magnitude of the torque is τ=pEsin⁡θ\tau = p E \sin \theta, where θ\theta is the angle between p⃗\vec{p} and E⃗\vec{E}.

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Stable Equilibrium: Occurs at θ=0∘\theta = 0^\circ. The potential energy is minimum (U=−pEU = -pE) and the torque is zero.

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Unstable Equilibrium: Occurs at θ=180∘\theta = 180^\circ. The potential energy is maximum (U=+pEU = +pE) and the torque is zero.

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Potential Energy: The work done in rotating the dipole from an angle θ1\theta_1 to θ2\theta_2 is stored as potential energy: U(θ)=−p⃗⋅E⃗=−pEcos⁡θU(\theta) = -\vec{p} \cdot \vec{E} = -pE \cos \theta.

📐Formulae

p⃗=q⋅2a⃗\vec{p} = q \cdot 2\vec{a}

τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}

τ=pEsin⁡θ\tau = pE \sin \theta

W=∫θ1θ2pEsin⁡θ dθ=pE(cos⁡θ1−cos⁡θ2)W = \int_{\theta_1}^{\theta_2} pE \sin \theta \, d\theta = pE(\cos \theta_1 - \cos \theta_2)

U=−pEcos⁡θU = -pE \cos \theta

💡Examples

Problem 1:

An electric dipole with dipole moment 4×10−9 C m4 \times 10^{-9} \text{ C m} is aligned at 30∘30^\circ with the direction of a uniform electric field of magnitude 5×104 N/C5 \times 10^4 \text{ N/C}. Calculate the magnitude of the torque acting on the dipole.

Solution:

Given: p=4×10−9 C mp = 4 \times 10^{-9} \text{ C m}, E=5×104 N/CE = 5 \times 10^4 \text{ N/C}, and θ=30∘\theta = 30^\circ. Using the formula τ=pEsin⁡θ\tau = pE \sin \theta: τ=(4×10−9)×(5×104)×sin⁡30∘\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ τ=20×10−5×12=10×10−5=10−4 N m\tau = 20 \times 10^{-5} \times \frac{1}{2} = 10 \times 10^{-5} = 10^{-4} \text{ N m}.

Explanation:

The torque is calculated by finding the product of the dipole moment, the electric field strength, and the sine of the angle between them.

Problem 2:

A dipole of moment p=6×10−8 C mp = 6 \times 10^{-8} \text{ C m} is rotated from stable equilibrium (θ=0∘θ = 0^\circ) to 60∘60^\circ in a field E=2×104 N/CE = 2 \times 10^{4} \text{ N/C}. Calculate the change in potential energy.

Solution:

U1=−pEcos⁡0∘=−(6×10−8)(2×104)(1)=−12×10−4 J=−0.0012 JU_1 = -pE \cos 0^\circ = -(6 \times 10^{-8})(2 \times 10^4)(1) = -12 \times 10^{-4} \text{ J} = -0.0012 \text{ J} U2=−pEcos⁡60∘=−(6×10−8)(2×104)(0.5)=−6×10−4 J=−0.0006 JU_2 = -pE \cos 60^\circ = -(6 \times 10^{-8})(2 \times 10^4)(0.5) = -6 \times 10^{-4} \text{ J} = -0.0006 \text{ J} Change in Potential Energy ΔU=U2−U1\Delta U = U_2 - U_1: −0.0006−(−0.0012)0.0006\begin{array}{r} -0.0006 \\ - (-0.0012) \\ \hline 0.0006 \end{array} Thus, ΔU=6×10−4 J\Delta U = 6 \times 10^{-4} \text{ J}.

Explanation:

Potential energy is calculated at both positions using U=−pEcos⁡θU = -pE \cos \theta. The change in potential energy is the difference between the final and initial states.