To find the torque experienced by the electric dipole when it is rotated by \( 30^\circ \), we first need to understand the relationship between torque (\( \tau \)), electric field (\( E \)), dipole moment (\( p \)), and the angle (\( \theta \)) between the dipole and the electric field.
The torque (\( \tau \)) experienced by a dipole of moment \( \mathbf{p} \) in a uniform electric field \( \mathbf{E} \) is given by the equation:
\(\tau = pE \sin \theta\)
Here, the dipole moment \( p \) is the product of the charge (\( q \)) and the separation distance (\( d \)):
\(p = q \times d\)
Given:
First, calculate the electric field (\( E \)) using the voltage and separation:
\(E = \frac{V}{s} = \frac{5}{0.5 \times 10^{-3}} = 10,000 \, V/m\)
Next, calculate the dipole moment:
\(p = q \times d = (2 \times 10^{-6}) \times (0.5 \times 10^{-6}) = 1 \times 10^{-12} \, Cm\)
Now, calculate the torque:
\(\tau = pE \sin \theta = (1 \times 10^{-12}) \times (10,000) \times \sin 30^\circ\)
Since \( \sin 30^\circ = 0.5 \):
\(\tau = (1 \times 10^{-12} \, Cm) \times (10,000 \, V/m) \times 0.5 = 5 \times 10^{-9} \, Nm\)
Therefore, the torque is \(5 \times 10^{-9} \, Nm\), which matches the correct answer option.
Step 1: Identify the given parameters and convert them to SI units.
Magnitude of each charge, \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \)
Separation distance between the charges (dipole length), \( d = 0.5 \, \mu m = 0.5 \times 10^{-6} \, m \)
Potential difference across the capacitor plates, \( V = 5 \, V \)
Separation between the capacitor plates, \( D = 0.5 \, mm = 0.5 \times 10^{-3} \, m \)
Angle by which the dipole is rotated from the electric field direction, \( \theta = 30^\circ \)
Step 2: Calculate the electric field between the capacitor plates.
The electric field \( E \) between the plates of a parallel plate capacitor is given by: \[ E = \frac{V}{D} \]
Substituting the given values: \[ E = \frac{5 \, V}{0.5 \times 10^{-3} \, m} = 10^4 \, V/m \]
Step 3: Calculate the dipole moment \( p \).
The dipole moment \( p \) is given by the product of the magnitude of one of the charges and the separation distance between the charges: \[ p = q \times d \] Substituting the given values: \[ p = (2 \times 10^{-6} \, C) \times (0.5 \times 10^{-6} \, m) = 1 \times 10^{-12} \, C \cdot m \]
Step 4: Calculate the torque \( \tau \) on the dipole.
The torque \( \tau \) on an electric dipole placed in a uniform electric field \( E \) at an angle \( \theta \) with the field is given by: \[ \tau = p E \sin \theta \] Substituting the calculated values: \[ \tau = (1 \times 10^{-12} \, C \cdot m) \times (10^4 \, V/m) \times \sin(30^\circ) \] We know that \( \sin(30^\circ) = \frac{1}{2} \). \[ \tau = 1 \times 10^{-8} \times \frac{1}{2} \, N \cdot m = 0.5 \times 10^{-8} \, N \cdot m = 5 \times 10^{-9} \, N \cdot m \] The value of the torque is \( 5 \times 10^{-9} \, Nm \), which corresponds to option (1).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)