CBSE 2024, Set 1

Question

An electric dipole consists of point charges $-1.0	ext{ pC}$ and $+1.0	ext{ pC}$ located at $(0, 0)$ and $(3	ext{ mm}, 4	ext{ mm})$ respectively in the $x-y$ plane. An electric field $\vec{E} = \left(1000 \frac{	ext{V}}{	ext{m}}ight)\hat{i}$ is switched on in the region. Find the torque $\vec{	au}$ on the dipole.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** Given: * Charge $q = 1.0	ext{ pC} = 1.0 	imes 10^{-12}	ext{ C}$ * Position of $-q$ is at the origin $(0, 0)$. * Position of $+q$ is at $(3	ext{ mm}, 4	ext{ mm})$. * Electric field $\vec{E} = 10^3\hat{i}	ext{ V/m}$. First, find the displacement vector $2\vec{a}$ connecting $-q$ to $+q$: $$2\vec{a} = (3 - 0)\hat{i} + (4 - 0)\hat{j} = (3\hat{i} + 4\hat{j})	ext{ mm}$$ $$2\vec{a} = (3\hat{i} + 4\hat{j}) 	imes 10^{-3}	ext{ m}$$ The dipole moment $\vec{p}$ is defined as $\vec{p} = q(2\vec{a})$: $$\vec{p} = (1.0 	imes 10^{-12}) 	imes (3\hat{i} + 4\hat{j}) 	imes 10^{-3}$$ $$\vec{p} = (3\hat{i} + 4\hat{j}) 	imes 10^{-15}	ext{ C}\cdot	ext{m}$$ The torque $\vec{	au}$ experienced by the dipole is given by $\vec{	au} = \vec{p} 	imes \vec{E}$: $$\vec{	au} = [(3\hat{i} + 4\hat{j}) 	imes 10^{-15}] 	imes [10^3\hat{i}]$$ $$\vec{	au} = 10^{-12} [(3\hat{i} 	imes \hat{i}) + (4\hat{j} 	imes \hat{i})]$$ Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$: $$\vec{	au} = 10^{-12} [0 + 4(-\hat{k})]$$ $$\vec{	au} = -4 	imes 10^{-12}\hat{k}	ext{ N}\cdot	ext{m}$$ The magnitude of the torque is $4 	imes 10^{-12}	ext{ N}\cdot	ext{m}$, and it is directed along the negative Z-axis.

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Physics > Electrostatics And Current Electricity > Electric Charges And Fields

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