Question 307: Electrostats & Capacitance

Question

A dipole with two electric charges of 2 µC magnitude each, with separation distance 0.5 µm, 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° from the axis, it tends to realign in the direction due to a torque. The value of torque is:

Accepted Answer

Given:
Potential difference $ V = 5 \, 	ext{V} $
Separation between the plates $ d = 0.5 \, 	ext{mm} = 0.5 	imes 10^{-3} \, 	ext{m} $
The electric field $ E $ is calculated as:
$ E = \frac{V}{d} = \frac{5}{0.5 	imes 10^{-3}} = 10^4 \, 	ext{V/m} $
The torque $ 	au $ experienced by the dipole in the electric field is given by:
$ 	au = PE \sin 	heta $
Where:
$ P $ is the dipole moment,
$ E $ is the electric field,
$ 	heta = 30^\circ $ is the angle of rotation.
The dipole moment $ P $ is calculated by:
$ P = q 	imes a = (2 	imes 10^{-6} \, 	ext{C}) 	imes (0.5 	imes 10^{-6} \, 	ext{m}) = 1 	imes 10^{-12} \, 	ext{C} \cdot 	ext{m} $
Substituting the values into the torque equation:
$ 	au = (1 	imes 10^{-12} \, 	ext{C} \cdot 	ext{m}) 	imes 10^4 \, 	ext{V/m} 	imes \sin 30^\circ  $
Since $\sin 30^\circ = 0.5$, the torque simplifies to:
$ 	au = 1 	imes 10^{-12} 	imes 10^4 	imes 0.5 = 5 	imes 10^{-9} \, 	ext{N} \cdot 	ext{m} $
Thus, the torque is $ 5 	imes 10^{-9} \, 	ext{N} \cdot 	ext{m} $.

Answer

2.5×10−9 Nm

Answer

2.5×10−12 Nm

Answer

5×10−3 Nm

Answer

5×10−9 Nm

Electrostats & Capacitance

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