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Electric Charges and Fields - Electric Dipole

Grade 12CBSEPhysics

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

πŸ”‘Concepts

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An electric dipole consists of a pair of equal and opposite point charges qq and βˆ’q-q, separated by a small distance 2a2a.

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The electric dipole moment pβƒ—\vec{p} is a vector quantity with magnitude p=qΓ—2ap = q \times 2a, directed from the negative charge βˆ’q-q to the positive charge +q+q.

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For a point far away from the dipole (r≫ar \gg a), the electric field EE decreases as 1/r31/r^3, which is faster than the 1/r21/r^2 dependence seen in a single point charge.

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In a uniform electric field E⃗\vec{E}, a dipole experiences a net torque τ⃗=p⃗×E⃗\vec{\tau} = \vec{p} \times \vec{E}, but the net translational force is zero.

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Stable equilibrium occurs when pβƒ—\vec{p} is parallel to Eβƒ—\vec{E} (ΞΈ=0∘\theta = 0^\circ), while unstable equilibrium occurs when pβƒ—\vec{p} is anti-parallel to Eβƒ—\vec{E} (ΞΈ=180∘\theta = 180^\circ).

πŸ“Formulae

p⃗=q⋅(2a⃗)\vec{p} = q \cdot (2\vec{a})

Eaxial=14πϡ02pr(r2βˆ’a2)2β‰ˆ14πϡ02pr3Β (forΒ r≫a)E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - a^2)^2} \approx \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \text{ (for } r \gg a)

Eequatorial=14πϡ0p(r2+a2)3/2β‰ˆ14πϡ0pr3Β (forΒ r≫a)E_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \text{ (for } r \gg a)

Ο„βƒ—=pβƒ—Γ—Eβƒ—=pEsin⁑θ\vec{\tau} = \vec{p} \times \vec{E} = pE \sin \theta

U=βˆ’pβƒ—β‹…Eβƒ—=βˆ’pEcos⁑θU = -\vec{p} \cdot \vec{E} = -pE \cos \theta

W=pE(cos⁑θ1βˆ’cos⁑θ2)W = pE(\cos \theta_1 - \cos \theta_2)

πŸ’‘Examples

Problem 1:

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

Solution:

Given: p=4Γ—10βˆ’9Β CΒ mp = 4 \times 10^{-9} \text{ C m}, E=5Γ—104Β N/CE = 5 \times 10^4 \text{ N/C}, ΞΈ=30∘\theta = 30^\circ. Using the formula Ο„=pEsin⁑θ\tau = pE \sin \theta: Ο„=(4Γ—10βˆ’9)Γ—(5Γ—104)Γ—sin⁑30∘\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ Ο„=20Γ—10βˆ’5Γ—0.5=10Γ—10βˆ’5Β NΒ m=10βˆ’4Β NΒ m\tau = 20 \times 10^{-5} \times 0.5 = 10 \times 10^{-5} \text{ N m} = 10^{-4} \text{ N m}

Explanation:

The torque is calculated by finding the cross product of the dipole moment and the electric field intensity.

Problem 2:

Calculate the potential energy difference when a dipole with pE=500pE = 500 units is rotated from θ=60∘\theta = 60^\circ to θ=90∘\theta = 90^\circ. Show the subtraction of final and initial energy values.

Solution:

Initial Potential Energy U1=βˆ’pEcos⁑60∘=βˆ’500Γ—0.5=βˆ’250U_1 = -pE \cos 60^\circ = -500 \times 0.5 = -250. Final Potential Energy U2=βˆ’pEcos⁑90∘=βˆ’500Γ—0=0U_2 = -pE \cos 90^\circ = -500 \times 0 = 0. To find Ξ”U=U2βˆ’U1\Delta U = U_2 - U_1: 0βˆ’(βˆ’250)250\begin{array}{r} 0 \\ - (-250) \\ \hline 250 \end{array} Ξ”U=250\Delta U = 250 units.

Explanation:

Potential energy is minimum when the dipole is aligned with the field. Rotating it requires work, which increases the potential energy of the system.