Electric Field Of Charged Disk - Practice Questions & MCQ

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)  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 () so, . The charge per unit area is  from this -  the charge of ring is  . The field component  at point P due to charge dQ of a ring of radius  r is - 

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

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

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

Special case - 

1) When , then         Note that, this equation is independent of 'x'

2)  When  (i.e very near to disc) , then