Question

Two identical positive charges are $0.6$ meters apart. Which is true about the electric field and the potential at the point which is directly in between the two?
A. The electric field and the potential are both equal to zero.
B. The electric field and the potential are both positive numbers.
C. The electric field is a positive number while the potential is a zero.
D. The electric field is zero while the potential is a positive number.

Answer 1

First of all, we will find the electric field due to the point charge and then we will find the electric potential. Then we will analyse the direction of the resultant electric field and resultant electric potential.

Complete step by step answer:
There are two identical positive charges which are placed in such a way that a distance of $0.6$ meters separates them.
We are asked to find a necessary condition which satisfies the given case.
To begin with, we are to find if there will be any changes in the electric field and the electric potential due to the positive charges or the identical charges.
First of all, we will find the electric field which originates because of the positive charge which is placed at certain distance, can be found out by the formula given below:
$E = \dfrac{{kq}}{{{r^2}}}$
Where,
$E$ indicates an electric field.
$k$ indicates proportionality constant.
$q$ indicates the magnitude of the charge.
$r$ indicates the distance.
The electric field is directed away from the positive charge. As we can see that both the charges are equal in magnitude and they are also same in terms of polarity, so the both the electric fields produced by these two charges are equal and opposite to each other. Hence, the two electric fields cancel out each other and the resultant becomes zero.
Again, we take the case of electric potential which is due to a point charge which is also positive in nature which is placed at a certain distance, can be found out by the formula given below:
$V = \dfrac{{kq}}{r}$
Where,
$V$ indicates the electric potential.
$k$ indicates proportionality constant.
$q$ indicates the magnitude of the charge.
$r$ indicates the distance.
Now, we find out the potential due to a point charge which is placed at a distance of $0.3\,{\text{m}}$ , which is given by:
$V' = \dfrac{{kq}}{{0.3}}$
Again, electric potential due to another point charge placed at a distance of $0.3\,{\text{m}}$ , which is given by:
$V'' = \dfrac{{kq}}{{0.3}}$
Now, we find out the total or resultant potential at the centre of a line joining the two identical charges is given by:
$V' + V'' \\\ \Rightarrow V’ + V” = \dfrac{{kq}}{{0.3}} + \dfrac{{kq}}{{0.3}} \\\ \Rightarrow V’ + V” = \dfrac{{2kq}}{{0.3}} \\\$
Hence, the electric potential is a non-zero positive number.

Hence, the correct option is D.

Note: While solving this problem remember that electric potential is scalar quantity which has no direction. It only has magnitude, so there is no point of cancelling out each other. However, on the other hand, the electric field is a vector quantity, which has direction, so here comes a point where the electric fields can cancel out each other, if they are opposite.