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Electric Field Of Charged Disk - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Electric field due to uniformly charged disc is considered one the most difficult concept.

  • 7 Questions around this concept.

Solve by difficulty

 A thin disc of radius b=2a has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge '\sigma 'on it. If the electric field on its axis at height' h ' (h<<a) from its centre is given as 'Ch' then value of 'C' is :

 

Concepts Covered - 1

Electric field due to uniformly charged disc

Electric field due to uniformly charged disk

Let us take a disk of radius R with a uniform positive surface charge density (charge per unit area) \sigma. Our aim is to find an electric field at a point on the axis of the disk at a distance x from its centre. 

                                                                          

 From the figure, we can see that the we have taken a  typical ring has charge dQ, inner radius r and outer radius r+d r . Its area d A
is approximately equal to its width d r times its circumference (2 \pi r) so, dA=2 \pi r .d r. The charge per unit area is \sigma=d Q / d A, from this -  the charge of ring is d Q=\sigma(2 \pi r d r) . The field component d E_{x} at point P due to charge dQ of a ring of radius  r is - 

                                                                 d E_{x}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \pi \sigma r d r) x}{\left(x^{2}+r^{2}\right)^{3 / 2}}

If we integrate from 0 to R, we will get the total field - 

                                          E_{x}=\int d E_{x}=\int_{0}^{R} d E_{x}=\int_{0}^{R} \frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \pi \sigma r d r) x}{\left(x^{2}+r^{2}\right)^{3 / 2}}

Here, 'x' is constant and 'r' is the variable. After integration we get - 

                                          E_{x}=\frac{\sigma x}{2 \varepsilon_{0}}\left[-\frac{1}{\sqrt{x^{2}+R^{2}}}+\frac{1}{x}\right]=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right]

 

As this disc is symmetric to x-axis, so the field in the rest of the component is zero i.e., E_y=E_z=0

 

Special case - 

1) When R>>xthen   E_{x}=\frac{\sigma}{2 \varepsilon_{0}}      Note that this equation is independent of 'x'

2)  When x\rightarrow 0 (i.e very near to disc), then E_{x}=\frac{\sigma}{2 \varepsilon_{0}}

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Electric field due to uniformly charged disc

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