
Find the electric field \[E\] at point \[P\] (as shown in the figure) on the perpendicular bisector of a uniformly charged thin wire of length \[L\] carrying a charge \[Q\]. The distance of the point \[P\] from the center of the rod is \[a = \left( {\frac{{\sqrt 3 }}{2}} \right)L\]

A. \[\frac{Q}{{2\sqrt 3 \pi {\varepsilon _o}{L^2}}}\]
B. \[\frac{{\sqrt 3 Q}}{{4\pi {\varepsilon _0}{L^2}}}\]
C. \[\frac{Q}{{3\pi {\varepsilon _o}{L^2}}}\]
D. \[\frac{Q}{{4\pi {\varepsilon _o}{L^2}}}\]
Answer
164.4k+ views
Hint: Write down the potential resulting from a point charge first. To determine the potential at point \[P\], integration on the line charge must be done. Therefore, choose a wire element \[\left( {dx} \right)\] and calculate its potential using the same methods as for a point charge.
Formula Used:
\[{{\rm{E}}_{{\rm{net }}}} = \frac{{{\rm{K}}\lambda }}{{\rm{r}}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)\]
Where \[{\theta _1},{\theta _2}\] are angle making with point \[{\rm{P}}\] from end.
Charge per unit length is \[\lambda \]
Complete answer:
We have been given in the question that,
Length of the wire \[ = L\]
Charge in the wire \[ = Q\]
Distance of the point \[P\] from the center of the rod is \[a = \left( {\frac{{\sqrt 3 }}{2}} \right)L\]

From the above diagram, we get
\[\tan \theta = \frac{{{\rm{ Perpendicular }}}}{{{\rm{ Base }}}}\]
Now, we have to observe the diagram and write the values in the above formula, we obtain\[\tan \theta = \frac{{\frac{L}{2}}}{{\frac{{\sqrt 3 }}{2}L}} = \frac{1}{{\sqrt 3 }}\]
On solving the above equation, we get
\[\theta = {30^\circ }\]
Now, we have to find electric field \[E\] at point \[P\]
Now,
\[{E_{net}} = \left( {\frac{{kQ}}{{\frac{{\sqrt 3 {L^2}}}{2}}}} \right)\left( {\sin {{30}^\circ } + \sin {{30}^\circ }} \right)\]
On solving the above equation by multiplying the terms inside the parentheses, we get
\[ = \left( {\frac{{2kQ}}{{\sqrt 3 {L^2}}}} \right)\left( {\frac{1}{2} + \frac{1}{2}} \right)\]
Now, we have to simplify the resultant equation, we have
\[ = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{2Q}}{{\sqrt 3 {L^2}}}} \right)\]
On canceling the similar terms in the above expression, we get
\[ = \frac{Q}{{2\sqrt 3 \pi {\varepsilon _0}{L^2}}}\]
Where: The net electric field is \[{\rm{E}}\].
The constant is \[{\rm{K}}\].
The charge in the wire is \[{\rm{Q}}\].
The wire's length is \[{\rm{L}}\].
Absolute permittivity is \[{\varepsilon _0}\].
Therefore, the electric field \[E\] at point \[P\] (as shown in the figure) on the perpendicular bisector of a uniformly charged thin wire of length \[L\] carrying a charge \[Q\] is \[\frac{Q}{{2\sqrt 3 \pi {\varepsilon _0}{L^2}}}\].
Hence, the option A is correct.
Note: Students are likely to make mistake in these types of problems. In a brief, it can be claimed that because electric potential is a scalar number, it is zero at all points along the perpendicular bisector. The position of the force on a positive test charge is assumed to equal the position of the field.
Formula Used:
\[{{\rm{E}}_{{\rm{net }}}} = \frac{{{\rm{K}}\lambda }}{{\rm{r}}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)\]
Where \[{\theta _1},{\theta _2}\] are angle making with point \[{\rm{P}}\] from end.
Charge per unit length is \[\lambda \]
Complete answer:
We have been given in the question that,
Length of the wire \[ = L\]
Charge in the wire \[ = Q\]
Distance of the point \[P\] from the center of the rod is \[a = \left( {\frac{{\sqrt 3 }}{2}} \right)L\]

From the above diagram, we get
\[\tan \theta = \frac{{{\rm{ Perpendicular }}}}{{{\rm{ Base }}}}\]
Now, we have to observe the diagram and write the values in the above formula, we obtain\[\tan \theta = \frac{{\frac{L}{2}}}{{\frac{{\sqrt 3 }}{2}L}} = \frac{1}{{\sqrt 3 }}\]
On solving the above equation, we get
\[\theta = {30^\circ }\]
Now, we have to find electric field \[E\] at point \[P\]
Now,
\[{E_{net}} = \left( {\frac{{kQ}}{{\frac{{\sqrt 3 {L^2}}}{2}}}} \right)\left( {\sin {{30}^\circ } + \sin {{30}^\circ }} \right)\]
On solving the above equation by multiplying the terms inside the parentheses, we get
\[ = \left( {\frac{{2kQ}}{{\sqrt 3 {L^2}}}} \right)\left( {\frac{1}{2} + \frac{1}{2}} \right)\]
Now, we have to simplify the resultant equation, we have
\[ = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{2Q}}{{\sqrt 3 {L^2}}}} \right)\]
On canceling the similar terms in the above expression, we get
\[ = \frac{Q}{{2\sqrt 3 \pi {\varepsilon _0}{L^2}}}\]
Where: The net electric field is \[{\rm{E}}\].
The constant is \[{\rm{K}}\].
The charge in the wire is \[{\rm{Q}}\].
The wire's length is \[{\rm{L}}\].
Absolute permittivity is \[{\varepsilon _0}\].
Therefore, the electric field \[E\] at point \[P\] (as shown in the figure) on the perpendicular bisector of a uniformly charged thin wire of length \[L\] carrying a charge \[Q\] is \[\frac{Q}{{2\sqrt 3 \pi {\varepsilon _0}{L^2}}}\].
Hence, the option A is correct.
Note: Students are likely to make mistake in these types of problems. In a brief, it can be claimed that because electric potential is a scalar number, it is zero at all points along the perpendicular bisector. The position of the force on a positive test charge is assumed to equal the position of the field.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: 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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
