
Two charges \[ + 3.2 \times {10^{ - 19}}C\] and \[ - 3.2 \times {10^{ - 19}}C\] placed at \[2.4\mathop A\limits^ \circ \] apart from an electric dipole. It is placed in a uniform electric field of intensity \[4 \times {10^5}V/m\]. The electric dipole moment is
A. \[15.36 \times {10^{ - 29}}Cm\]
B. \[15.36 \times {10^{ - 19}}Cm\]
C. \[7.68 \times {10^{ - 29}}Cm\]
D. \[7.68 \times {10^{ - 19}}Cm\]
Answer
161.7k+ views
Hint:A dipole is the combination of two charges of equal in magnitude and opposite in nature. The electric dipole moment is proportional to the magnitude of the charge and the separation between the charges.
Formula used:
\[p = qd\]
Here p is the magnitude of the dipole moment, q is the magnitude of charge separated at distance d.
Complete step by step solution:
It is given that two charges are \[ + 3.2 \times {10^{ - 19}}C\] and \[ - 3.2 \times {10^{ - 19}}C\].
\[{Q_1} = + 3.2 \times {10^{ - 19}}C\]
\[\Rightarrow {Q_2} = - 3.2 \times {10^{ - 19}}C\]
The separation between the charges is \[2.4\mathop A\limits^ \circ \]. As we know that the angstrom is the unit of distance.One angstrom is equal to \[{10^{ - 10}}m\]. We need to convert the given distance unit into S.I. unit.
So, the distance between the charges in S.I. unit is,
\[d = 2.4 \times {10^{ - 10}}m\]
The magnitude of the charge is equal to \[3.2 \times {10^{ - 19}}C\].
\[q = 3.2 \times {10^{ - 19}}C\]
Using the formula to find the magnitude of the electric dipole moment,
\[p = qd\]
Here, p is the magnitude of the electric dipole moment, q is the magnitude of the charge separated by the distance d.
The magnitude of the dipole moment is,
\[p = \left( {3.2 \times {{10}^{ - 19}}C} \right) \times \left( {2.4 \times {{10}^{ - 10}}} \right)Cm\]
\[\therefore p = 7.68 \times {10^{ - 29}}Cm\]
Hence, the magnitude of the electric dipole moment is \[7.68 \times {10^{ - 19}}Cm\]. The electric dipole moment is a vector quantity, so the electric dipole moment is represented using the magnitude of the electric dipole moment as well the direction of the electric dipole moment.
Therefore, the correct option is C.
Note: As the electric field lines due to negative point charge is directed towards the charge and the electric field lines due to positive point charge is directed away from the charge. So the direction of the electric dipole moment is assumed to be from positive charge to the negative charge.
Formula used:
\[p = qd\]
Here p is the magnitude of the dipole moment, q is the magnitude of charge separated at distance d.
Complete step by step solution:
It is given that two charges are \[ + 3.2 \times {10^{ - 19}}C\] and \[ - 3.2 \times {10^{ - 19}}C\].
\[{Q_1} = + 3.2 \times {10^{ - 19}}C\]
\[\Rightarrow {Q_2} = - 3.2 \times {10^{ - 19}}C\]
The separation between the charges is \[2.4\mathop A\limits^ \circ \]. As we know that the angstrom is the unit of distance.One angstrom is equal to \[{10^{ - 10}}m\]. We need to convert the given distance unit into S.I. unit.
So, the distance between the charges in S.I. unit is,
\[d = 2.4 \times {10^{ - 10}}m\]
The magnitude of the charge is equal to \[3.2 \times {10^{ - 19}}C\].
\[q = 3.2 \times {10^{ - 19}}C\]
Using the formula to find the magnitude of the electric dipole moment,
\[p = qd\]
Here, p is the magnitude of the electric dipole moment, q is the magnitude of the charge separated by the distance d.
The magnitude of the dipole moment is,
\[p = \left( {3.2 \times {{10}^{ - 19}}C} \right) \times \left( {2.4 \times {{10}^{ - 10}}} \right)Cm\]
\[\therefore p = 7.68 \times {10^{ - 29}}Cm\]
Hence, the magnitude of the electric dipole moment is \[7.68 \times {10^{ - 19}}Cm\]. The electric dipole moment is a vector quantity, so the electric dipole moment is represented using the magnitude of the electric dipole moment as well the direction of the electric dipole moment.
Therefore, the correct option is C.
Note: As the electric field lines due to negative point charge is directed towards the charge and the electric field lines due to positive point charge is directed away from the charge. So the direction of the electric dipole moment is assumed to be from positive charge to the negative charge.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Wheatstone Bridge for JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
