Question

When the separation between two charges is increased, the electric potential energy of the charges
$A.$ Increases
$B.$ Decreases
$C.$ Remains the same
$D.$ May increase or decrease

Answer 1

To solve this problem properly we should apply the basic relation of distance between charges and electric potential energy. Nature of the charges is also considered to analyse the change in electric potential energy of the charges. Electric potential energy is defined as the energy which is needed to move a charge against an electric field.

Complete step by step answer:
We know that the potential energy between two point charges is inversely proportional to the distance between them.
Mathematically we can write, ${{U}_{E}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{r}$,
Where, ${{U}_{E}}=$ Electric potential energy, $k=$ coulomb constant, ${{q}_{1}}$ and ${{q}_{2}}$ are charges and $r$ is the distance between the charges.
On the basis of our above discussion we come on the conclusion that if the separation between both the charges is increased then the electric potential energy decreases.

Hence, option $(B)$ is correct.

Additional Information:
We should know about the basics of charge. Charge is defined as the basic property of a body by virtue of which it experiences a force when placed in an electromagnetic field. SI unit of charge is Coulomb $(C)$. Charge is of two types, positive charge and negative charge. Basic law of charge is that charges repel each other and unlike charges attract each other.

Note:
In solving these types of problems we should consider the signs of the charges involved. Relation of electric potential energy is also considered. We are only dealing with point charges and not the combination of charges. We should also remember that If both the charges have the same sign then potential energy will decrease with increase in separation and if both the charges have opposite signs then the electric potential energy will increase with increase of distance between them.