Question

Two positive point charges q are located on the y-axis at$y=\pm \frac{1}{2}s$.

a.Find an expression for the potential along the x-axis.

b.Draw a graph of V versus x for $-\infty <x<\infty$. For comparison, use a dotted line to show the potential of a point charge 2qlocated at the origin.

Short answer

(a) The expression for potential along$x-axis$is$V\left(x\right)=\frac{4kq}{\sqrt{4x^2+s^2}}$.

(b) The graph is given as

Step-by-step solution

Part (a) Step 1: Given Information

We have given point charge located at$y=\pm \frac{1}{2}s$.

Part (a) Step 2: Explanation

The potential created by the two charges by symmetry will be twice the potential created by one of the charges. The potential as a function of the$x-$coordinate will be the potential at a distance of

localid="1648325162527" $d=\sqrt{x^2+{(s/2)}^2}=\frac{1}{2}\sqrt{4x^2+s^2}$

So, it is said that the potential will be

$$\begin{gathered} V\left(x\right)=2k\frac{q}{d} \\ =2k\frac{q}{\left(1/2\right)\sqrt{4x^2+s^2}} \\ V\left(x\right)=\frac{4kq}{\sqrt{4x^2+s^2}} \end{gathered}$$

Part (b) Step 1: Given Information

We need to draw a graph of $V$vesus$x$for$-\infty <x<\infty$.

Part (b) Step 2: Explanation

The potential of a charge $2q$at the center of the coordinate system is given as

$V\left(x\right)=\frac{2kq}{x}$

Now, we can say the two potentials would be the same only if$s=0$. For any value of except zero, the potential coming from the separated charges will be lower than that of the double charge at the center.

And also the behavior of the two functions at the$x\to 0$limiting case must to notice. In this case the potential for the split charges reaches a maximum. On the other side, if we have a charge at the center, the potential will tend to infinity.

So, we can say that the two potentials are the same in the$x\to \infty$limiting case.

Finally, we come up to the diagram