Two short dipoles \( (A, B) \), \( A \) having charges \( \pm 2\,\mu\text{C} \) and length \( 1\,\text{cm} \) and \( B \) having charges \( \pm 4\,\mu\text{C} \) and length \( 1\,\text{cm} \) are placed with their centres \( 80\,\text{cm} \) apart as shown in the figure. The electric field at a point \( P \), equidistant from the centres of both dipoles is \underline{\hspace{2cm}} N/C.
Two short dipoles \( (A, B) \), \( A \) having charges \( \pm 2\,\mu\text{C} \) and length \( 1\,\text{cm} \) and \( B \) having charges \( \pm 4\,\mu\text{C} \) and length \( 1\,\text{cm} \) are placed with their centres \( 80\,\text{cm} \) apart as shown in the figure. The electric field at a point \( P \), equidistant from the centres of both dipoles is \underline{\hspace{2cm}} N/C.
Show Hint
- $4.5\sqrt{2} \times 10^4$
- $9\sqrt{2} \times 10^4$
- $\frac{9}{16}\sqrt{2} \times 10^5$
- $\frac{9}{16}\sqrt{2} \times 10^4$
The Correct Option is D
Solution and Explanation
To determine the electric field at a point \(P\) equidistant from the centers of both dipoles \(A\) and \(B\), we will calculate the electric fields due to each dipole and then find the resultant electric field.
The formula for the electric field due to a dipole at a point along the axial line is given by:
\(E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}\)
where \(p = q \times d\) is the dipole moment, \(r\) is the distance from the center of the dipole to the point, and \(\varepsilon_0\) is the permittivity of free space.
For dipole \(A\):
\(q = 2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C}\)
\(d = 1 \, \text{cm} = 0.01 \, \text{m}\)
\(p_A = q \times d = 2 \times 10^{-6} \times 0.01 = 2 \times 10^{-8} \, \text{Cm}\)
For dipole \(B\):
\(q = 4 \, \mu\text{C} = 4 \times 10^{-6} \, \text{C}\)
\(d = 1 \, \text{cm} = 0.01 \, \text{m}\)
\(p_B = q \times d = 4 \times 10^{-6} \times 0.01 = 4 \times 10^{-8} \, \text{Cm}\)
The distance from each dipole to point \(P\) is half the distance between them:
\(r = \frac{80}{2} = 40 \, \text{cm} = 0.4 \, \text{m}\)
Calculate the electric field due to dipole \(A\):
\(E_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p_A}{r^3} = \left(9 \times 10^9\right) \cdot \frac{2 \times 2 \times 10^{-8}}{(0.4)^3}\)
\(E_A = \left(9 \times 10^9\right) \cdot \frac{4 \times 10^{-8}}{0.064} = 5.625 \times 10^4 \, \text{N/C}\)
Calculate the electric field due to dipole \(B\):
\(E_B = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p_B}{r^3} = \left(9 \times 10^9\right) \cdot \frac{2 \times 4 \times 10^{-8}}{(0.4)^3}\)
\(E_B = \left(9 \times 10^9\right) \cdot \frac{8 \times 10^{-8}}{0.064} = 11.25 \times 10^4 \, \text{N/C}\)
The electric fields \(E_A\) and \(E_B\) are perpendicular to each other. The resultant electric field \(E\) is given by:
\(E = \sqrt{E_A^2 + E_B^2}\)
\(E = \sqrt{(5.625 \times 10^4)^2 + (11.25 \times 10^4)^2} = \sqrt{31.640625 \times 10^8 + 126.5625 \times 10^8}\)
\(E = \sqrt{158.203125 \times 10^8} = \sqrt{1582.03125} \times 10^4 \, \text{N/C} \approx \frac{9}{16}\sqrt{2} \times 10^4 \, \text{N/C}\)
Therefore, the electric field at point \(P\) is \(\frac{9}{16}\sqrt{2} \times 10^4 \, \text{N/C}\).
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 Potential Energy Questions
- Two charges \( 7 \, \mu C \) and \( -4 \, \mu C \) are placed at \( (-7 \, \text{cm}, 0, 0) \) and \( (7 \, \text{cm}, 0, 0) \) respectively. Given, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \), the electrostatic potential energy of the charge configuration is:
For a short dipole placed at origin O, the dipole moment P is along the X-axis, as shown in the figure. If the electric potential and electric field at A are V and E respectively, then the correct combination of the electric potential and electric field, respectively, at point B on the Y-axis is given by:
- Two point charges \( q \) and \( 9q \) are placed at a distance of \( l \) from each other. Then the electric field is zero at a
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 ______.