Question

Three charges, each equal to $q$ , are placed at the three corners of a square a side $a$, find the electric field at the fourth corner
A. $\left( {\sqrt 2 + 1} \right)\dfrac{q}{{8\pi {\varepsilon _0}{a^2}}}$
B. $\left( {2\sqrt 2 + 1} \right)\dfrac{q}{{8\pi {\varepsilon _0}{a^2}}}$
C. $\left( {2\sqrt 2 + 1} \right)\dfrac{q}{{4\pi {\varepsilon _0}{a^2}}}$
D. $0$

Answer 1

We will calculate the electric field at the point D by calculating the net magnetic of the electric fields ${\vec E_1}$ , ${\vec E_2}$ and ${\vec E_3}$. Firstly we will calculate the magnitude of the vectors of electric fields ${E_1}$ and ${E_2}$, and then we will add it to the electric field of ${\vec E_3}$.

Complete step by step answer:
Here, we will consider a square whose each side is $a$and a charge $q$ is placed on the three corners of the square.

Here, charge $q$ is placed at the corners A, B and C. As we know that a positive charge induces an electric field away from it. Therefore, ${E_1}$ is the electric field induced by the charge placed at C, ${E_2}$ is the electric field induced by the charge placed at A and ${E_3}$ is the charge induced by the charge placed at B.

Now, electric field induced by the charge at C,
${E_1} = \dfrac{q}{{4\pi {\varepsilon _0}{a^2}}}$
Where $K$ is the proportionality constant, $q$ is the charge and $a$ is the side.
Also, electric field induced by charge placed at A,
${E_2} = \dfrac{q}{{4\pi {\varepsilon _0}{a^2}}}$
And, electric field induced by charge placed at B,
${E_3} = \dfrac{q}{{8\pi {\varepsilon _0}{a^2}}}$
Here, the area in case of ${E_3}$ is taken as $\sqrt 2 a$ because the charge is making a diagonal with the electric field.

Now, the net electric field at point D is given by
${E_{net}} = \left( {{{\vec E}_1} + {{\vec E}_2}} \right) + {E_3}$
Here, $\left( {{{\vec E}_1} + {{\vec E}_2}} \right)$ means that we are taking the magnitude of the vectors of electric fields ${E_1}$ and ${E_2}$ and is shown below
$\left| {{{\vec E}_1} + {{\vec E}_2}} \right| = \sqrt {{{\left( {{E_1} + {E_2}} \right)}^2}}$
$\Rightarrow \,{\vec E_1} + {\vec E_2} = \sqrt {E_1^2 + E_2^2 + 2{E_1}{E_2}\cos \theta }$
Here, ${E_1}$ and ${E_2}$ are the same, therefore, ${E_1} = {E_2} = E$ .

Also, here $\theta = 90^\circ$ because the angle made by the electric fields ${E_1}$ and ${E_2}$ is $90^\circ$.
Therefore, the above equation becomes
${\vec E_1} + {\vec E_2} = \sqrt {{E^2} + {E^2} + 2E.E\cos 90^\circ }$
$\Rightarrow {\vec E_1} + {\vec E_2} = \sqrt {2{E^2}}$
$\Rightarrow \,{\vec E_1} + {\vec E_2} = \sqrt 2 E$
$\Rightarrow \,{\vec E_1} + {\vec E_2} = \sqrt 2 \dfrac{q}{{4\pi {\varepsilon _0}{a^2}}}$
Therefore, putting this value in ${E_{net}}$ , we get
${E_{net}} = \sqrt 2 \dfrac{q}{{4\pi {\varepsilon _0}{a^2}}} + \dfrac{q}{{8\pi {\varepsilon _0}{a^2}}}$
$\Rightarrow \,{E_{net}} = \dfrac{q}{{\pi {\varepsilon _0}}}\left( {\dfrac{{\sqrt 2 }}{{4{a^2}}} + \dfrac{1}{{8{a^2}}}} \right)$
$\Rightarrow \,{E_{net}} = \dfrac{q}{{\pi {\varepsilon _0}}}\left( {\dfrac{{2\sqrt 2 + 1}}{{8{a^2}}}} \right)$
$\therefore E = \left( {2\sqrt 2 + 1} \right)\dfrac{q}{{8\pi {\varepsilon _0}{a^2}}}$
Hence, the electric field at the forth corner D is $\left( {2\sqrt 2 + 1} \right)\dfrac{q}{{8\pi {\varepsilon _0}{a^2}}}$.

Hence, option B is the correct option.

Note: As we know that the positive charge will induce the electric field away from it, therefore, the charge $q$ placed at C will induce the electric field in the direction of the diagonal of the square. Therefore, in the case of the electric field ${\vec E_3}$ , we will take the distance as $\sqrt 2 a$.