At which distance along the centre axis of a uniformly charged plastic disc of radius R is the magnitude of the electric field equal to one-half the magnitude of the field at the centre of the surface of the disc ?

Options

A. `R/(sqrt(2))`
B. `R/(sqrt(3))`
C. `sqrt(2)R`
D. `sqrt(3R)`

Answer

B

Explanation

At a point on the axis of uniformly charged disc at a distance x above the centre of the disc, the magnitude of the electric field is
`E=(sigma)/(2 epsilon_(0))[1-x/(sqrt(x^(2)+R^(2)))]`
but `E_(c)=(sigma)/(2epsilon_(0))` such that `E/(E_(c))=1/2`
Then `1-x/(sqrt(x^(2)+R^(2)))=1/2` or `x/(sqrt(x^(2)+R^(2)))=1/2`
Squaring both sides and multiplying by `x^(2)+R^(2)` to obtain
`x^(2)-(x^(2))/4+(R^(2))/4`
Thus `x^(2)=(R^(2))/3l,x=R/(sqrt(3))`