Question

A right isosceles triangle of side $a$ has charges $q, + 3q$ and $- q$ arranged on its vertices as shown in figure. What is the electric potential at $P$ midway between the line connecting the $+ q$ and $- q$ charges?

A. $\dfrac{{3q}}{{r{\varepsilon _0}a}}$
B. $\dfrac{{3q}}{{\sqrt 2 \pi {\varepsilon _0}a}}$
C. $\dfrac{q}{{\pi {\varepsilon _0}a}}$
D. $\dfrac{{3q}}{{2\sqrt 2 \pi {\varepsilon _0}a}}$

Answer 1

Use the superposition principle, to find the electric potential at a point in a system with multiple charges. Use the known geometrical relationships to find the distances between the charges and the point.

Complete step by step answer:
Let us first label the points in the diagram, to refer to them easily while using geometrical relationships.

We are asked to find the electric potential at the point P due to the charges $+ q, - q{\text{ and }} + 3q$. We will first find out the potential at P due to a single charge and then use the superposition principle to find the collective electric potential at P. And for this we need to know the distances $AP,BP,CP$.

We are given that $\vartriangle ABC$ is a right-angled triangle and point P is the midpoint of $AC$. So, by using Pythagoras theorem we have, $AC = \sqrt 2 a$ and $AP = PC = \dfrac{a}{{\sqrt 2 }}$. We can also observe that $\vartriangle ABP \cong \vartriangle CBP$ [by SSS congruence criterion]. And so, $\angle ABP = \angle CBP = 45^\circ$. Also, since $\vartriangle ABC$ is isosceles we have $\angle A = \angle C = 45^\circ$. Now since, sides opposite to opposite angles are equal, in $\vartriangle ABP$, we have $AP = BP = \dfrac{a}{{\sqrt 2 }}$ [because $\angle ABP = \angle BAP = 45^\circ$].

So, we now know the distances between the charges and the point P;
$AP = BP = CP = \dfrac{a}{{\sqrt 2 }}$
So, the electric potential at P due to $+ q$ is,
${V_1} = \dfrac{q}{{4\pi {\varepsilon _0}\dfrac{a}{{\sqrt 2 }}}} = \dfrac{q}{{2\sqrt 2 \pi {\varepsilon _0}a}}$
Electric potential at P due to $- q$ is,
${V_2} = \dfrac{{ - q}}{{4\pi {\varepsilon _0}\dfrac{a}{{\sqrt 2 }}}} = \dfrac{{ - q}}{{2\sqrt 2 \pi {\varepsilon _0}a}}$
And, electric potential at P due to $+ q$ is,
${V_3} = \dfrac{{3q}}{{4\pi {\varepsilon _0}\dfrac{a}{{\sqrt 2 }}}} \\\ \Rightarrow {V_3}= \dfrac{{3q}}{{2\sqrt 2 \pi {\varepsilon _0}a}}$
We know from the superposition of electric potential that the electric potential of a system of charges is the sum of potentials of the individual charges.
So, the electric potential at P $V = {V_1} + {V_2} + {V_3}$.
$V = \dfrac{q}{{2\sqrt 2 \pi {\varepsilon _0}a}} + \dfrac{{ - q}}{{2\sqrt 2 \pi {\varepsilon _0}a}} + \dfrac{{3q}}{{2\sqrt 2 \pi {\varepsilon _0}a}}$
$\therefore V = \dfrac{{3q}}{{2\sqrt 2 \pi {\varepsilon _0}a}}$

Hence, the answer is option D.

Note: The electric potential at a point is the amount of work which is needed to move a charge from a reference point to that point. Usually, we consider the reference point at infinity. Whereas, potential difference between two points is the amount of work that needs to be done to bring a charge from one point to the other.