(A) The electric field $E$ between the plates of the capacitor is given by $E = \frac{V}{d}$,where $V = 5 \ V$ and $d = 0.5 \ mm = 5 \times 10^{-4} \

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(A) The electric field $E$ between the plates of the capacitor is given by $E = \frac{V}{d}$,where $V = 5 \ V$ and $d = 0.5 \ mm = 5 \times 10^{-4} \ m$.$E = \frac{5}{5 \times 10^{-4}} = 10^4 \ V/m$.The dipole moment $p$ is given by $p = q \times a$,where $q = 2 \ \mu C = 2 \times 10^{-6} \ C$ and $a = 0.5 \ \mu m = 5 \times 10^{-7} \ m$.$p = 2 \times 10^{-6} \times 5 \times 10^{-7} = 1 \times 10^{-12} \ C \cdot m$.The torque $\tau$ acting on the dipole is given by $\tau = pE \sin \theta$,where $\theta = 30^{\circ}$.$\tau = (1 \times 10^{-12} \ C \cdot m) \times (10^4 \ V/m) \times \sin(30^{\circ})$.$\tau = 10^{-8} \times 0.5 = 5 \times 10^{-9} \ N \cdot m$.

Class 12 Physics Electric Charges and Fields Electric Dipole and Electric Field

Medium

$A$ dipole with two electric charges of $2 \ \mu C$ magnitude each,with separation distance $0.5 \ \mu m$,is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of $5 \ V$ is applied. Separation between the plates is $0.5 \ mm$. If the dipole is rotated by $30^{\circ}$ from the axis,it tends to realign in the direction due to a torque. The value of torque is $:$

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A
$5 \times 10^{-9} \ Nm$
B
$5 \times 10^{-3} \ Nm$
C
$2.5 \times 10^{-12} \ Nm$
D
$2.5 \times 10^{-9} \ Nm$

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Electric Charges and Fields

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Electric Dipole and Electric Field

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Three point charges $q, -2q$ and $q$ are placed along the $x$-axis at $x = -a, 0$ and $a$ respectively. As $a \rightarrow 0$ and $q \rightarrow \infty$ while $qa^2 = Q$ remains finite, the electric field at a point $P$, at a distance $x$ $(x \gg a)$ from $x = 0$ is $E = \frac{\alpha Q}{4 \pi \epsilon_0 x^\beta} \hat{i}$. Then:

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$A$ molecule with a dipole moment $p$ is placed in an electric field of strength $E$. Initially,the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti-parallel to the field,the work required to be done by an external agent is:

Two charges of $+25 \times 10^{-9} \, C$ and $-25 \times 10^{-9} \, C$ are placed $6 \, m$ apart. Find the ratio of the electric field intensity at a point $4 \, m$ from the center of the electric dipole on the $(i)$ axial line and $(ii)$ equatorial line.

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