Question

In the above figure, what is the net electric potential at point P due to the four particles if $V = 0$ at infinity, $q = 5.00fC$ and $d = 4.00cm$ ?

Answer 1

Electric field is a vector quantity, while electric potential is a scalar quantity. Thus, to find the electric potential on a point due to several other point charges, we could make a sum of their individual electric potential with respect to the point charge. So, to find the electric potential on point P due to the other point charges, we can find the electric potential of each point charge on P and add them, $V = {V_1} + {V_2} + {V_3} + {V_4}$ .

Formulas used We will be using the formula to find the electric potential of a point charge on a body given by, $V = \dfrac{{kq}}{d}$ , where $V$ is the electric potential on a point by a point charge, $q$ is the magnitude of the point charge, $d$ is the distance between the charge and the point, and $k$ is the electrostatic constant given by, $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ .

Complete Step by Step Solution
We can see that there are 4 charges surrounding the point P, in 3 directions. There are 2 point charges $+ q$ at a distance $d$ in two opposite sides of point P, and a charge $- q$ at a distance $2d$ from P, and another charge $- q$ at a distance $d$ from the point P in the same direction along the previously mentioned charge.
The electric potential of charges at either side of point P on P are given by,
${V_1} = {V_2} = \dfrac{{k( + q)}}{d}$
Similarly, the electric potential on point P, due to the third charge,
${V_3} = \dfrac{{k( - q)}}{{2d}}$
And the electric potential on point P due to the fourth charge will be,
${V_4} = \dfrac{{k( - q)}}{d}$
We know that electric potential is a scalar quantity, thus the total electric potential on point P is simply the sum of all the electric potentials on point P, $V = {V_1} + {V_2} + {V_3} + {V_4}$ .
${V_{total}} = \dfrac{{k( + q)}}{d} + \dfrac{{k( + q)}}{d} + \dfrac{{k( - q)}}{{2d}} + \dfrac{{k( - q)}}{d}$
${V_{total}} = kq\left[ {\dfrac{{( + 1)}}{d} + \dfrac{{( + 1)}}{d} + \dfrac{{( - 1)}}{{2d}} + \dfrac{{( - 1)}}{d}} \right]$
Substituting the value of $k = \dfrac{1}{{4\pi {\varepsilon _0}}}$ and solving,
${V_{total}} = \dfrac{1}{{4\pi {\varepsilon _0}}}q\left[ {\dfrac{{(2 + 2 - 1 - 2)}}{{2d}}} \right]$
${V_{total}} = \dfrac{1}{{4\pi {\varepsilon _0}}}q\left[ {\dfrac{{(1)}}{{2d}}} \right]$
We also know that the $q = 5.00fC = 5 \times {10^{ - 15}}C$ , $d = 4.00cm = 4 \times {10^{ - 2}}m$ and the value of the electrostatic constant, $k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 8.996 \times {10^9}N{m^2}{C^{ - 2}}$ .
${V_{total}} = 8.996 \times {10^9}N{m^2}{C^{ - 2}} \times 5 \times {10^{ - 15}}C\left[ {\dfrac{{(1)}}{{2 \times 4 \times {{10}^{ - 2}}m}}} \right]$
Solving for ${V_{total}}$ we get,
${V_{total}} = 5.5625 \times {10^{ - 4}}Nm/C$ .

Note
The electric potential is a scalar quantity because it is a ratio of the potential energy of the system per unit charge, but the electric field is a vector because it is the ratio of the electric force applied to the distance.