Two equal and opposite charges separated by a distance, d, experience an attractive force, F. If each charge is doubled in magnitude, what should be the new distance between them so that the force remains unchanged?

If the each charge is doubled in magnitude and force remains unchanged then the distance between the charged particle is four times the original distance.

Explanation:

The force between two charged objects are shown by Coulomb's law.

In the scalar form, this law is represented as:

where,

k = coulomb's constant

q₁ and q₂ are the magnitude of charged objects

d = distance between the charges.

According to the question,

Case1:

charge on object 1 = +q₁

Charge on object 2 = -q₂

Distance between the charges = d

Case 2:

Charge on object 1 = +2q₁

Charge on object 2 = -2q₂

Force between the objects is same

New distance, dₓ = ?

If both the forces are equal then,

On simplifying the equation we get,

So, if the each charge is doubled in magnitude and force remains unchanged then the distance between the charged particle is four times the original distance.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

found in the nucleus

has mass of one amu