Two point chrges of `30muCand40muC` are 20 cm apart from each other .Where will be the electric field zero on the line joining the from `30muC` charge ?

Options

A. `(20sqrt(3))/(sqrt(2)+1)` cm
B. `(20sqrt(3))/(sqrt3+2)cm`
C. `(20sqrt(3))/(sqrt(5))cm`
D. `5 cm`

Answer

B

Explanation

We can see that the horizontal field component from the upper and lower quadrants cancel and downwards components and thus , the field is pointed downwards . The calculation requires us to integrate the `1E_(y)` components along the upper quadrants .This yield the integral ,
`E_(y)=2int(kdQsintheta)/(R^(2))`
where , `theta` = angle pf points of the from Q runs from 0 to `(pi)/(2)` .
Here , `dQ=lambdaRd theta=(2Q)/(pid theta)`
and linear charge density , `lambda=(Q)/(piR//2)`
This calulation yields , `E_(y)=(4kQ)/(piR^(2))=(Q)/(pi^(2)epsilon_(0)R^(2))`