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Question: A point charge \(q\) is located at the center of a thin ring of radius \(R\) with uniformly distribu...

A point charge qq is located at the center of a thin ring of radius RR with uniformly distributed charge q-q. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance xx from its center if x>>Rx>>R.

Explanation

Solution

The net electric field strength at the point lying on the axis of the ring at a distance xx is equal to the sum of electric field strengths at that point due to the point charge qq at the center of the ring, as well as the ring of uniformly distributed charge q-q, itself. Since x>>Rx>>R, binomial approximation can be used while simplifying this net electric field strength.
Formula used:
1)Ec=kqx21){{E}_{c}}=\dfrac{kq}{{{x}^{2}}}
2)Er=kqx(x2+R2)322){{E}_{r}}=\dfrac{-kqx}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}
3)EP=Ec+Er3){{E}_{P}}={{E}_{c}}+{{E}_{r}}

Complete answer:
We are provided with a point charge qq, located at the center of a thin ring of radius RR with uniformly distributed charge q-q. We are required to find the magnitude of the electric field strength vector at a point lying on the axis of the ring at a distance xx from its center, if x>>Rx>>R.

From the diagram given above, it is clear that we have to find the magnitude of electric field strength at point PP. Also, the center of the ring where a charge qq is located is denoted as OO. The distance between center of the ring OO and the point on the axis of the ring PP is given by
OP=xOP=x
and radius of the ring OROR is given by
OR=ROR=R
Now, if we draw a right triangle ORPORP, then
RP=(R2+x2)12RP={{({{R}^{2}}+{{x}^{2}})}^{\dfrac{1}{2}}}
where
RPRP is the distance of the ring from the point lying on the axis of the ring
The net electric field strength at the point (P)(P), lying on the axis of the ring at a distance xx is equal to the sum of electric field strengths at that point due to the point charge qq at the center of the ring as well as the ring of uniformly distributed charge q-q, itself.
If Ec{{E}_{c}} denotes the electric field string at PP due to charge qqat the center of the ring, then, Ec{{E}_{c}} is given by
Ec=kqx2{{E}_{c}}=\dfrac{kq}{{{x}^{2}}}
where
Ec{{E}_{c}} is the electric field string at point PP due to charge qq
k=14πε0k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}, is the electrostatic constant
qq is the point charge on the center of the ring OO
xx is the distance between the center of the ring and the point PP, lying on the axis of the ring (=OP)(=OP)
Let this be equation 1.
Similarly, if Er{{E}_{r}} represents the electric field strength at PP due to the ring of uniformly distributed charge qq, then, Er{{E}_{r}} is given by
Er=k(q)x(x2+R2)32=kqx(x2+R2)32{{E}_{r}}=\dfrac{k(-q)x}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{-kqx}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}
where
Er{{E}_{r}} is the electric field string at point PP due to the ring of uniformly distributed charge q-q
k=14πε0k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}, is the electrostatic constant
xx is the distance between the center of the ring and the point PP, lying on the axis of the ring (=OP)(=OP)
RR is the radius of the ring (=OR)(=OR)
Let this be equation 2.
Now, using equation 1 and equation 2, we have
EP=Ec+Er=kqx2kqx(x2+R2)32{{E}_{P}}={{E}_{c}}+{{E}_{r}}=\dfrac{kq}{{{x}^{2}}}-\dfrac{kqx}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}
where
EP{{E}_{P}} is the net electric field strength at point PP
Let this be equation 3.
Simplifying equation 3, we have
EP=kqx2kqx(x2+R2)32=kqx3(xx(1+(Rx)2)32)=kqx2(1(1+(Rx)2)32){{E}_{P}}=\dfrac{kq}{{{x}^{2}}}-\dfrac{kqx}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{kq}{{{x}^{3}}}\left( x-\dfrac{x}{{{\left( 1+{{\left( \dfrac{R}{x} \right)}^{2}} \right)}^{\dfrac{3}{2}}}} \right)=\dfrac{kq}{{{x}^{2}}}\left( 1-{{\left( 1+{{\left( \dfrac{R}{x} \right)}^{2}} \right)}^{-\dfrac{3}{2}}} \right)
Now, we are given that x>>Rx>>R. Therefore, we have
x>>RRx<<1x>>R\Rightarrow \dfrac{R}{x}<<1
Hence, we can use the binomial approximation given by
(1+a)n=1na{{(1+a)}^{-n}}=1-na for a>1a>1
in the above equation.
Using this binomial approximation in the above simplification of EP{{E}_{P}}, we have
EP=kqx2(1(1+(Rx)2)32)=kqx2(1(132(Rx)2))=kqx2(11+32(Rx)2)=3kqR22x4{{E}_{P}}=\dfrac{kq}{{{x}^{2}}}\left( 1-{{\left( 1+{{\left( \dfrac{R}{x} \right)}^{2}} \right)}^{-\dfrac{3}{2}}} \right)=\dfrac{kq}{{{x}^{2}}}\left( 1-\left( 1-\dfrac{3}{2}{{\left( \dfrac{R}{x} \right)}^{2}} \right) \right)=\dfrac{kq}{{{x}^{2}}}\left( 1-1+\dfrac{3}{2}{{\left( \dfrac{R}{x} \right)}^{2}} \right)=\dfrac{3kq{{R}^{2}}}{2{{x}^{4}}}
Let this be equation 4.
Substituting the expression for electrostatic constant k=14πε0k=\dfrac{1}{4\pi {{\varepsilon }_{0}}} in equation 4, we have
EP=3kqR22x4=3qR28πε0x4{{E}_{P}}=\dfrac{3kq{{R}^{2}}}{2{{x}^{4}}}=\dfrac{3q{{R}^{2}}}{8\pi {{\varepsilon }_{0}}{{x}^{4}}}
Therefore, if x>>Rx>>R, the magnitude of electric field strength vector at a point lying on the axis of the ring ( of uniformly distributed charge q-q), at a distance xx from its center (of charge qq) is equal to 3qR28πε0x4\dfrac{3q{{R}^{2}}}{8\pi {{\varepsilon }_{0}}{{x}^{4}}}.

Note:
Students need to be aware of the formula for electric field strength due to a ring or a coil, at a point on the axis of the ring or the coil. As already mentioned, it is given by
E=kqx(x2+R2)32E=\dfrac{kqx}{{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\dfrac{3}{2}}}}
where
EE is the electric field strength due to a ring carrying a charge qq at a point on the axis of the ring
xx is the distance of the point on the axis of the ring from the center of the ring
RR is the radius of the ring
k=14πε0k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}, is the electrostatic constant
This is a standard definition and can be easily derived by considering a small charge element (dq)(dq)on the ring and deducing the electric field strength due to this charge element at a point on the axis of the ring. Integrating this field strength over the circumference of the ring will give the expression for field strength due to the whole ring at this point on its axis, as given above.