Question

Six point charges are placed at the vertices of a hexagon of side $1m$ as shown in figure. Net electric field at the centre of the hexagon is?

Answer 1

Here, you are given a hexagon and at each corner of the hexagon, you have a point charge. You are asked to find the electric field at the centre of the hexagon due to all the charges. In order to solve this question, you need to first find the electric field due to a single point charge at some point of interest. After knowing the electric field expression, use that to find the electric due to each point charge at the centre of the hexagon and finally use the superposition theorem to find the net electric field.

Complete step by step answer:
Let us first find the electric field due to a point charge. Consider a point charge having charge equal to $q$ located at the origin. Another point charge having charge equal to $q'$ is located at some position vector $\overrightarrow r$. It is observed that the charge $q$ exerts a force on the charge $q'$. This force is given by Coulomb’s law and the expression of force is given as $\overrightarrow F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qq'}}{{{r^2}}}\widehat r$.

Electric field is defined as force per unit charge. Here, the charge $q'$ is a test charge. Mathematically, the electric field is given as $\overrightarrow E = \dfrac{{\overrightarrow F }}{{q'}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\widehat r$. The magnitude of electric field is $\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$.Now, let us come back to our question. Starting from the leftmost charge $q$ and letting it be number 1, we will go anti-clockwise. The electric field due to each charge at the centre of the hexagon will be as follows:

\Rightarrow{E_2} = \dfrac{q}{{4\pi {\varepsilon _0}{d^2}}}\left( {\dfrac{1}{2}\widehat i + \dfrac{{\sqrt 3 }}{2}\widehat j} \right) \\\ \Rightarrow{E_3} = \dfrac{q}{{4\pi {\varepsilon _0}{d^2}}}\left( { - \dfrac{1}{2}\widehat i + \dfrac{{\sqrt 3 }}{2}\widehat j} \right) \\\ \Rightarrow{E_4} = \dfrac{{ - q}}{{4\pi {\varepsilon _0}{d^2}}}\left( { - \widehat i} \right) \\\ \Rightarrow{E_5} = \dfrac{{ - q}}{{4\pi {\varepsilon _0}{d^2}}}\left( { - \dfrac{1}{2}\widehat i - \dfrac{{\sqrt 3 }}{2}\widehat j} \right) \\\ \Rightarrow{E_6} = \dfrac{{ - q}}{{4\pi {\varepsilon _0}{d^2}}}\left( {\dfrac{1}{2}\widehat i - \dfrac{{\sqrt 3 }}{2}\widehat j} \right) \\\ $$ According to superposition theorem, the net electric field at the centre of the hexagon will be equal to the vector sum of all the electric fields. $E = {E_1} + {E_2} + {E_3} + {E_4} + {E_5} + {E_6} \\\ \Rightarrow E = \dfrac{q}{{4\pi {\varepsilon _0}{d^2}}}\left( {\widehat i} \right) + \dfrac{q}{{4\pi {\varepsilon _0}{d^2}}}\left( {\dfrac{1}{2}\widehat i + \dfrac{{\sqrt 3 }}{2}\widehat j} \right) + \dfrac{q}{{4\pi {\varepsilon _0}{d^2}}}\left( { - \dfrac{1}{2}\widehat i + \dfrac{{\sqrt 3 }}{2}\widehat j} \right) + \dfrac{{ - q}}{{4\pi {\varepsilon _0}{d^2}}}\left( { - \widehat i} \right) + \dfrac{{ - q}}{{4\pi {\varepsilon _0}{d^2}}}\left( { - \dfrac{1}{2}\widehat i - \dfrac{{\sqrt 3 }}{2}\widehat j} \right) + \dfrac{{ - q}}{{4\pi {\varepsilon _0}{d^2}}}\left( {\dfrac{1}{2}\widehat i - \dfrac{{\sqrt 3 }}{2}\widehat j} \right) \\\ \Rightarrow E = \dfrac{q}{{4\pi {\varepsilon _0}{d^2}}}\left( {2\widehat i + 2\sqrt 3 \widehat j} \right) \\\ \Rightarrow E = \dfrac{q}{{4\pi {\varepsilon _0}{{\left( 1 \right)}^2}}}\left( 2 \right)\left( {\widehat i + \sqrt 3 \widehat j} \right) \\\ \therefore E = \dfrac{q}{{2\pi {\varepsilon _0}}}\left( {\widehat i + \sqrt 3 \widehat j} \right) \\\ $ **Therefore, net electric field at the centre of the hexagon is $\dfrac{q}{{2\pi {\varepsilon _0}}}\left( {\widehat i + \sqrt 3 \widehat j} \right)$.** **Note:** We have discussed in detail about the electric field at any point due to a point charge, we derived this expression using Coulomb's law, so keep this in mind. Also, while finding electric fields, write properly the unit vector and also carefully keep track of the sign of charge, whether it is not negative or positive.