Two point charges \( -4 \, \mu C \) and \( 4 \, \mu C \), constituting an electric dipole, are placed at \( (-9, 0, 0) \, \text{cm} \) and \( (9, 0, 0) \, \text{cm} \) in a uniform electric field of strength \( 10^4 \, \text{N/C} \). The work done on the dipole in rotating it from the equilibrium through \( 180^\circ \) is:
Show Hint
- 14.4 mJ
- 18.4 mJ
- 12.4 mJ
- 16.4 mJ
The Correct Option is A
Approach Solution - 1
The question involves calculating the work done on an electric dipole when it is rotated from its equilibrium position through \( 180^\circ \) in a uniform electric field. The key concepts here are the electric dipole moment and the potential energy in an external electric field.
Conceptual Approach:
An electric dipole consists of two equal and opposite charges separated by a distance. When placed in a uniform electric field, the dipole experiences torque which tries to align it along the field direction. The work done on the dipole when rotated in the electric field is given by changing its potential energy.
Formulas Required:
- The dipole moment \( \mathbf{p} \) is given by: \(\mathbf{p} = q \cdot \mathbf{d}\), where \( q \) is the charge and \( \mathbf{d} \) is the distance vector between the charges.
- The potential energy of a dipole in an electric field \( \mathbf{E} \) is: \(U = -\mathbf{p} \cdot \mathbf{E} = -pE \cos \theta\), where \( \theta \) is the angle between \( \mathbf{p} \) and \( \mathbf{E} \).
- Work done \( W \) when rotating the dipole from \( \theta_1 \) to \( \theta_2 \) is: \(W = U(\theta_2) - U(\theta_1)\).
Calculation:
- Calculate the dipole moment \( \mathbf{p} \):
The charges are \( -4 \, \mu \text{C} \) and \( 4 \, \mu \text{C} \), and the distance between them is from \( (-9, 0, 0) \) to \( (9, 0, 0) \), which is 18 cm or 0.18 m.
\(p = q \cdot d = 4 \times 10^{-6} \, \text{C} \times 0.18 \, \text{m} = 7.2 \times 10^{-7} \, \text{Cm}\) - Calculate Initial Potential Energy \( U(\theta_1 = 0^\circ) \):
\(U(0^\circ) = -pE \cos(0^\circ) = -7.2 \times 10^{-7} \times 10^4 \times 1 = -7.2 \times 10^{-3} \, \text{J}\) - Calculate Final Potential Energy \( U(\theta_2 = 180^\circ) \):
\(U(180^\circ) = -pE \cos(180^\circ) = -7.2 \times 10^{-7} \times 10^4 \times (-1) = 7.2 \times 10^{-3} \, \text{J}\) - Calculate the Work Done:
\(W = U(180^\circ) - U(0^\circ) = 7.2 \times 10^{-3} - (-7.2 \times 10^{-3}) = 14.4 \times 10^{-3} \, \text{J} = 14.4 \, \text{mJ}\)
Therefore, the work done on the dipole when rotated by \( 180^\circ \) is 14.4 mJ.
Approach Solution -2
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
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,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Electrostatics and Work Done by Electric Field Questions
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?
- Two metal plates (A, B) are kept horizontally with separation of \( \frac{12}{\pi} \) cm, with plate A on the top.} An atomizer jet sprays oil (density \( 1.5 \, \text{g/cm}^3 \)) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials \( V_A \) and \( V_B \) are required on plates A and B respectively in order to ensure the droplets do not descend. The values of \( V_A \) and \( V_B \) are _______. (Neglect the air resistance to the droplets and take \( g = 10 \, \text{m/s}^2 \))}
- The electric potential as a function of \(x,y\) is given by \(V=5(x^2-y^2)\,\text{V}\). The electric field at the point \((2,3)\) m is _____ V/m.
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
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 : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.