
The linear charge density of the circumference of a circle of radius $a$ varies as $\lambda = {\lambda _0}\cos \theta $. The total charge on it is …………………
A) $Zero$
B) $Infinite$
C) $\pi a{\lambda _0}$
D) $2\pi a$
Answer
159.3k+ views
Hint: To find the solution of this question, first of all we need to integrate the equation of linear charge density. After that we need to put the given value of $\lambda $and solve the equation. The solution of the equation will give the value of total charge on it. Also, we need to find the value of the angle $\theta $by which the radius of the circle is varying.
Complete step by step solution:
We know that linear charge density is given by, $\lambda = \dfrac{{dq}}{{dx}}$ …………………(i)
Where, we know that $\lambda $ is the charge density.
Also, $dq$ is the small amount of charge present on the circumference of the circle,
$dx$ is the circumferential length on which charge is distributed
Equation (i) can be written as,
$dq = \lambda dx$
Now, we need to integrate the above equation.
$\smallint dq = \smallint _0^{2\pi }\lambda dx$
$ \Rightarrow q = \smallint _0^{2\pi }{\lambda _0}\cos \theta dx$…………(ii)
We already know that, $\theta = \dfrac{{arc}}{{radius}}$
$ \Rightarrow d\theta = \dfrac{{dx}}{a}$
Or, $dx = ad\theta $
Now, putting the value of $dx$in equation (ii), we get,
$q = {\lambda _0}\smallint _0^{2\pi }\cos \theta .ad\theta $
$ \Rightarrow q = {\lambda _0}a\smallint _0^{2\pi }\cos \theta .d\theta $
$ \Rightarrow q = {\lambda _{_0}}a[\sin \theta ]_0^{2\pi }$
$ \Rightarrow q = {\lambda _0}a[\sin 2\pi - \sin 0]$
$ \Rightarrow q = {\lambda _0}a[0 - 0]$ $[\because \sin 2\pi = o]$
$ \Rightarrow q = {\lambda _0}a[0]$
$\therefore q = 0$
Hence, option (A), i.e. zero is the correct solution for the question.
Note: As, in the given question, we have to find the value of total charge, when linear charge density is given. Linear charge density is the quantity of charge per unit length and is measured in coulombs per meter, at any point on a line charge distribution. Charge can either be positive or negative, since electric charge can be either positive or negative. One should always be clear with the formula of linear charge density. We should not be confused with the surface charge density or volumetric charge density.
Surface charge density, $\sigma = qA$
Volumetric charge density, $\rho = \dfrac{q}{V}$.
Complete step by step solution:
We know that linear charge density is given by, $\lambda = \dfrac{{dq}}{{dx}}$ …………………(i)
Where, we know that $\lambda $ is the charge density.
Also, $dq$ is the small amount of charge present on the circumference of the circle,
$dx$ is the circumferential length on which charge is distributed
Equation (i) can be written as,
$dq = \lambda dx$
Now, we need to integrate the above equation.
$\smallint dq = \smallint _0^{2\pi }\lambda dx$
$ \Rightarrow q = \smallint _0^{2\pi }{\lambda _0}\cos \theta dx$…………(ii)
We already know that, $\theta = \dfrac{{arc}}{{radius}}$
$ \Rightarrow d\theta = \dfrac{{dx}}{a}$
Or, $dx = ad\theta $
Now, putting the value of $dx$in equation (ii), we get,
$q = {\lambda _0}\smallint _0^{2\pi }\cos \theta .ad\theta $
$ \Rightarrow q = {\lambda _0}a\smallint _0^{2\pi }\cos \theta .d\theta $
$ \Rightarrow q = {\lambda _{_0}}a[\sin \theta ]_0^{2\pi }$
$ \Rightarrow q = {\lambda _0}a[\sin 2\pi - \sin 0]$
$ \Rightarrow q = {\lambda _0}a[0 - 0]$ $[\because \sin 2\pi = o]$
$ \Rightarrow q = {\lambda _0}a[0]$
$\therefore q = 0$
Hence, option (A), i.e. zero is the correct solution for the question.
Note: As, in the given question, we have to find the value of total charge, when linear charge density is given. Linear charge density is the quantity of charge per unit length and is measured in coulombs per meter, at any point on a line charge distribution. Charge can either be positive or negative, since electric charge can be either positive or negative. One should always be clear with the formula of linear charge density. We should not be confused with the surface charge density or volumetric charge density.
Surface charge density, $\sigma = qA$
Volumetric charge density, $\rho = \dfrac{q}{V}$.
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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
