
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
219k+ 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
A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

ill in the blanks Pure tungsten has A Low resistivity class 12 physics JEE_Main

The value of the resistor RS needed in the DC voltage class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

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

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

