Answer
Verified
110.4k+ 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
Write an article on the need and importance of sports class 10 english JEE_Main
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
If x2 hx 21 0x2 3hx + 35 0h 0 has a common root then class 10 maths JEE_Main
The radius of a sector is 12 cm and the angle is 120circ class 10 maths JEE_Main
For what value of x function fleft x right x4 4x3 + class 10 maths JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
In Searles apparatus when the experimental wire is class 11 physics JEE_Main
The energy stored is a condenser is in the form of class 12 physics JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main