Answer:
here north are not vector option b hope ur help
The electrical force acting on a charge q immersed in an electric field is equal to

where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
Answer:
Explanation:
In case of diffraction , angular width of central maxima =2 λ/d
λ is wave length of light and d is slit width
In case of interference , angular width of each fringe
= λ /D
D is distance between two slits
No of interference fringe in central diffraction fringe
=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.
Benthos
Option b is the answer