Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

A point charge

q

is located at the centre of a thin ring of the radius

R

, with uniformly distributed charge

- q

. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring, at a distance

x

from its centre, if

x ≫ R

.

Accepted Answer

Problem statement: The given problem is based on the calculation of electric field intensity using superposition of electric potential due to a uniform thin ring and a charge.

The relation between electric potential and electric field intensity is given as,

$E = - \frac{d V}{d x}$ , where, $x$ is the distance.

Solution:

The electric potential at a distance $x$ from the given ring, is given by,

$V (x) = V_{charge} + V_{ring} = \frac{q}{4 π ε_{0} x} - \frac{q}{4 π ε_{0} {(R^{2} + x^{2})}^{\frac{1}{2}}}$

Hence, the field strength along $x$ -axis, (which is the net field strength in our case) is,

$E_{x} = |\frac{d V}{d x}| = \frac{q}{4 π ε_{0}} \frac{1}{x^{2}} - \frac{q x}{4 π ε_{0} {(R^{2} + x^{2})}^{\frac{3}{2}}}$

$= \frac{\frac{q}{4 π ε_{0}} x^{3} [{(1 + \frac{R^{2}}{x^{2}})}^{\frac{3}{2}} - 1]}{x^{2} {(R^{2} + x^{2})}^{\frac{3}{2}}}$

$= \frac{\frac{q}{4 π ε_{0}} x^{3} [(1 + \frac{3}{2} \frac{R^{2}}{x^{2}} + \frac{3}{8} \frac{R^{4}}{x^{4}} + \dots) - 1]}{x^{2} {(R^{2} + x^{2})}^{\frac{3}{2}}}$

Neglecting the higher power of $\frac{R}{x}$ , as $x ≫ R$ .

$= \frac{q}{4 π ε_{0}} \frac{x^{3} (\frac{3 R^{2}}{2 x^{2}})}{x^{5} {[1 + {(\frac{R}{x})}^{2}]}^{\frac{3}{2}}} = \frac{q}{4 π ε_{0}} \times \frac{3 R^{2}}{2 x^{4}} {[1 + {(\frac{R}{x})}^{2}]}^{- \frac{3}{2}}$

$= \frac{q}{4 π ε_{0}} \times \frac{3 R^{2} [1 - \frac{3 R^{2}}{2 x^{2}}]}{2 x^{4}}$

$\Rightarrow E \approx \frac{q}{4 π ε_{0}} \times \frac{3 R^{2}}{2 x^{4}} = \frac{3 q R^{2}}{8 π ε_{0} x^{4}}$

A point charge $q$ is located at the centre of a thin ring of the radius $R$ , with uniformly distributed charge $- q$ . Find the magnitude of the electric field strength vector at the point lying on the axis of the ring, at a distance $x$ from its centre, if $x ≫ R$ .

Important Questions on ELECTRODYNAMICS

Learn and Prepare for any exam you want !

A point charge q is located at the centre of a thin ring of the radius R, with uniformly distributed charge -q. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring, at a distance x from its centre, if x≫R.

Important Questions on ELECTRODYNAMICS

Learn and Prepare for any exam you want !

A point charge $q$ is located at the centre of a thin ring of the radius $R$ , with uniformly distributed charge $- q$ . Find the magnitude of the electric field strength vector at the point lying on the axis of the ring, at a distance $x$ from its centre, if $x ≫ R$ .