In Fig. 24-38, what is the net electric potential at point Pdue to the four particles if V = 0at infinity,q = 5.00 fC, and d = 4.00 cm?

Electric potential at infinity, V = 0

The given charge value, $q=5.00fC=5\times {10}^{-15}C$

The distance value, d = 4 cm = 0.04 m

The amount of work done to move a unit electric charge from a reference point to the specific point in the electric field is called the electric potential of the charge. It can also be sad that the potential is directly proportional to the charge and inversely proportional to the distance the charge is moved. Using this concept, the net potential of the system can be calculated.

Formula:

The electric potential at a point due to a charge,

$V=\frac{kq}{r}$ ….. (i)

Here, V is the electric potential, k is the Coulomb’s constant having a value $9\times {10}^{9}N·m^2/C^2,q$is the charge, and r is the distance.

The position of the charges from the field point P are shown in the figure below.

A charge -q is a distance 2d from P, a charge -q is a distance d from P, and two charges +q are each a distance d from P.

Thus, the net electric potential at point P can be given using equation (i) as follows:

$\begin{array}{rcl}V& =& kq\left(\frac{-1}{2d}-\frac{1}{d}+\frac{1}{d}+\frac{1}{d}\right)\\ & =& \frac{kq}{2d}\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}V& =& \frac{\left(9\times {10}^{9}N·m^2/C^2\right)\left(5\times {10}^{-15}C\right)}{2\left(0.04m\right)}\\ & =& 5.62\times {10}^{-4}V\end{array}$

Hence, the value of the net potential is $5.62\times {10}^{-4}V$.