Question

A charge $Q$ is spread uniformly in the form of a line charge density $\lambda = \dfrac{Q}{{3a}}$ on the sides of an equilateral triangle of perimeter $3a$. Calculate the potential at the centroid $C$ of the triangle.

Answer 1

To solve the question, you need to first draw the diagram of the given situation and then look at the fact that the potential at the centroid due to each of the halves of the sides of the equilateral triangle is equal. Hence, you can simply find the potential on the centroid due to a single half and then multiply it by six since there are six halves of the three sides of the triangle.

Complete step by step answer:
We will proceed with the solution exactly as explained in the hint section of the solution to the asked question. Firstly, we need to draw the diagram of the given situation, which can be drawn as follows:

As we can see, the potential at the centroid $C$ of the given triangle is symmetrical along the bisectors of the sides of the triangle and thus, to solve the question, we can simply find the potential due to any of the halves of the sides, and then multiply its value with six to reach at the final answer.
Let us first have a look at how to find the potential at the centroid due to the half of the base as shown in the diagram:
The line charge density has been given as $\lambda = \dfrac{Q}{{3a}}$
If we consider a small element with a thickness of $dx$ at a distance of $x$ from the perpendicular bisector of the base of the triangle, its small charge can be given as:
$dq = \lambda dx \\\ dq = \dfrac{Q}{{3a}}dx \\\$
As shown in the diagram, the distance between this small element and the centroid of the triangle is going to be $r$ which can also be written as:
$r = \sqrt {{x^2} + {{\left( {\dfrac{a}{{2\sqrt 3 }}} \right)}^2}}$
Now, let us define the small potential at the centroid due to the small element above-mentioned as:
$dv = k\dfrac{{dq}}{r}$
Where, $k$ is the Coulomb’s constant which is also given as $k = \dfrac{1}{{4\pi {\varepsilon _o}}}$ and is numerically equal to $9 \times {10^9}\,N{m^2}/{C^2}$
$dq$ is the small charge of the small element and,
$r$ is the distance between the element and the centroid of the triangle
The net potential due to the half of the base can be given as:
$V = \int\limits_{x = 0}^{x = a/2} {dv}$
Substituting the value of $dv$, we get:
$V = \int\limits_{x = 0}^{x = a/2} {k\dfrac{{dq}}{r}}$
Now, if we substitute the values of $dq$ and $r$ as we talked above, we can write the equation as:
$V = k\int\limits_0^{a/2} {\dfrac{Q}{{3a}}\left( {\dfrac{{dx}}{{\sqrt {{x^2} + {{\left( {\dfrac{a}{{2\sqrt 3 }}} \right)}^2}} }}} \right)}$
We can see that the value of $Q,\,a$ is constant, hence, this can also be written as:
$V = k\dfrac{Q}{{3a}}\int\limits_0^{a/2} {\left( {\dfrac{{dx}}{{\sqrt {{x^2} + {{\left( {\dfrac{a}{{2\sqrt 3 }}} \right)}^2}} }}} \right)}$
Now, to integrate, we can use the integration identity as follows:
$\int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = \log \left( {x + \sqrt {{x^2} + {a^2}} } \right)}$
After applying this identity and putting in the values, we get the value of potential due to the half of the base of the triangle as:
$V = k\dfrac{Q}{{3a}}\log \left( {2 - \sqrt 3 } \right)$
Now, the final answer is not found yet, to find the final answer, we need to multiply this value with six as follows:
Answer $= \,6V = 6k\dfrac{Q}{{3a}}\log \left( {2 - \sqrt 3 } \right)$
Which can be simplified as:
Answer $= 2.634k\dfrac{Q}{a}$
So, this is the final answer to the asked question.

Note: Many students leave the solution at the point of finding the potential at the centroid due to half of the side of the triangle and forget to multiply it by six to find the correct answer and thus lose marks, do not make such mistakes while solving such questions.