Question

The force applied on 1 coulomb charge is 2 units, then electric field is​

Answer 1

Answer:

ans we know that the formula

F = q E

1 = 2 E

E = .5volt

Answer 2

Answer:

The electric field intensity is 2N/C.

Explanation:

Concept:

• The electric field at a location is the field that exerts force on a unit positive test charge placed in its vicinity.

• The relation between the electric field, $\vec E$ and the electric force, $\vec F$ on an arbitrary charge, ${q}$ is given by

                                           $\vec E = \dfrac{\vec F}{q}$

• Electric field measured in newtons/coulomb, acts in the same direction as the electric force.

Given the force applied on charge, $F= 2~ units$

And the charge, $q=1 coulomb$

Then the electric field intensity is

,$\vec E = \dfrac{\vec F}{q}== \dfrac{2}{1}=2N/C$

Therefore, the electric field intensity is 2N/C.