Acharge of +0.08 c moves to the right due to a 4 n force exerted by an electric field. what is the magnitude and direction of the electric field? a) 0.32 n/c left b) 0.32 n/c right c) 50 n/c left d) 50 n/c right

This is late but the answer is 50 n/c right

Answer : Electric field is given by, E = 50 N/C in right.

Explanation :

It is given that,

Charge, q = 0.08 C (moves to the right)

Force exerted by the electric field, F = 4 N

We know that the electric field is defined as :

The direction of field is same as the direction of electric force i.e.

in right direction.

So, the correct option is (D).

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.