Consider a positive charge q and a point b twice as far away from q as point a. what is the ratio of the electric field strength at point a to the electric field strength at point b

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

.

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

.

By substitution we get that

Now, letting be the electric field at point A, letting be the electric field at point B, and letting R be the distance from the charge to A:

.

The ration of the electric fields is

This means that at half the distance, the electric field is four times stronger.

the answer would be a

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards