Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

A thin non-conducting ring of radius $R$ has a linear charge density $λ = λ_{0} cosφ$ , where $λ_{0}$ is a constant, $φ$ is the azimuthal angle. Find the magnitude of the electric field strength,

(a) at the centre of the ring,

(b) on the axis of the ring as a function of the distance $x$ from its centre. Investigate the obtained function at $x ≫ R$ .

Accepted Answer

It is given that the ring has linear charge density, $λ = λ_{0} \cos φ$ .
Since $\cos φ$ is positive in first and fourth quadrants and negative in second and third quadrants, the charge distribution will look like the diagram given below. Hence, according to the above diagram, left half will have positive charge and right half will have negative charge and due to symmetry, the net electric field will be towards $(-) z$ -axis.

Let us take an element, an angle $φ$ from $(+) z$ -axis in clockwise direction which makes angle $d φ$ on the centre $O$ . Due to this element, the electric field will be in opposite direction of small element. Due to symmetry, all $\sin$ components of this small electric field $d E$ will cancel will each other and $\cos$ components will add to provide the net electric field in negative $z$ -direction.

$E_{net} = \int d E_{Z} = \int d E \cos φ$
$E_{net} = \int (\frac{k d q}{R^{2}}) \cos φ$ .

Elemental charge, $(d q) = λ d l$ ,
$\Rightarrow d q = λ_{0} \cos φ (R d φ)$ .

Put this value in the equation of $E_{net}$ ,

$E_{net} = \int_{0}^{2 π} (\frac{k λ_{0} \cos φ (R d φ)}{R^{2}}) \cos φ E_{net} = \int_{0}^{2 π} \frac{k λ_{0} \cos^{2} φ}{R} d φ E_{net} = \frac{k λ_{0}}{R} \int_{0}^{2 π} \cos^{2} φ d φ . Here, k = \frac{1}{4 π ε_{0}} E_{net} = \frac{λ_{0}}{4 {πε}_{0} R} \int_{0}^{2 π} \cos^{2} φ d φ E_{net} = \frac{λ_{0}}{2 ε_{0} R} \{∵ \int_{0}^{2 π} \cos^{2} φ d φ = π\}$

(b) Now, we have to find electric field at the axis of the ring as a function of the distance $x$ from its centre.

Take an element at an azimuthal angle $φ$ from the $x$ -axis, the element subtending an angle $d φ$ at the centre. The elementary field at $P$ due to the element is given by,
$d E = \frac{1}{4 π ε_{0}} \frac{d q}{{(\sqrt{R^{2} + x^{2}})}^{2}}$ .

From symmetry, we can see that the components of $d E$ due to different elements will add up along the direction parallel to $x -$ axis. The net electric field due to components parallel to $x -$ axis at point $P$ ,
${(E_{net})}_{1} = \int d E \cos α \{Here, \cos α = \frac{x}{\sqrt{R^{2} + x^{2}}}\} {(E_{net})}_{1} = \int \frac{1}{4 π ε_{0}} \frac{d q}{{(\sqrt{R^{2} + x^{2}})}^{2}} \cos α {(E_{net})}_{1} = \int_{0}^{2 π} (\frac{1}{4 π ε_{0}} \frac{λ_{0} \cos φ (d l)}{{(\sqrt{R^{2} + x^{2}})}^{2}}) \cos α {(E_{net})}_{1} = \int_{0}^{2 π} (\frac{1}{4 π ε_{0}} \frac{λ_{0} \cos φ (R d φ)}{{(\sqrt{R^{2} + x^{2}})}^{2}}) \frac{x}{\sqrt{R^{2} + x^{2}}} {(E_{net})}_{1} = \frac{λ_{0} x R}{4 π ε_{0} {(R^{2} + x^{2})}^{\frac{3}{2}}} \int_{0}^{2 π} \cos φ (d φ) {(E_{net})}_{1} = 0 \{∵ \int_{0}^{2 π} \cos φ (d φ)\}$

$d E \sin α$ will have two components: one along $y -$ axis and one along $z -$ axis at point $P$ . The components along $y -$ axis will cancel each other but those along $z -$ axis will add up at point $P$ . The net field due to this at point $P$ will be given by,

${(E_{net})}_{2} = = \int d E \sin α \cos φ \{Here, \sin α = \frac{R}{\sqrt{R^{2} + x^{2}}}\} {(E_{net})}_{2} = \int \frac{1}{4 π ε_{0}} \frac{d q \cos φ}{{(\sqrt{R^{2} + x^{2}})}^{2}} \frac{R}{\sqrt{R^{2} + x^{2}}} {(E_{net})}_{2} = \int_{0}^{2 π} (\frac{1}{4 π ε_{0}} \frac{λ_{0} \cos^{2} φ (d l)}{{(\sqrt{R^{2} + x^{2}})}^{2}}) \frac{R}{\sqrt{R^{2} + x^{2}}} {(E_{net})}_{2} = \int_{0}^{2 π} (\frac{1}{4 π ε_{0}} \frac{λ_{0} \cos^{2} φ (R^{2} d φ)}{{(R^{2} + x^{2})}^{\frac{3}{2}}}) = \frac{λ_{0} R^{2}}{4 π ε_{0} {(R^{2} + x^{2})}^{\frac{3}{2}}} \int_{0}^{2 π} \cos^{2} φ (d φ) {(E_{net})}_{2} = \frac{π λ_{0} R^{2}}{4 π ε_{0} {(R^{2} + x^{2})}^{\frac{3}{2}}} \{∵ \int_{0}^{2 π} \cos^{2} φ (d φ) = π\} {(E_{net})}_{2} = \frac{λ_{0} R^{2}}{4 ε_{0} {(R^{2} + x^{2})}^{\frac{3}{2}}}$

The net field at point $P$ ,

$\Rightarrow E_{net} = {(E_{net})}_{1} + {(E_{net})}_{2} \Rightarrow E_{net} = \frac{λ_{0} R^{2}}{4 ε_{0} {(R^{2} + x^{2})}^{\frac{3}{2}}} For, x > > R, \Rightarrow E_{net} = \frac{λ_{0} R^{2}}{4 ε_{0} x^{3}} = \frac{λ_{0} π R^{2}}{4 π ε_{0} x^{3}} \Rightarrow E_{net} = \frac{p}{4 π ε_{0} x^{3}} \{where, p = λ_{0} π R^{2}\}$

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