Four like charges each of +q are placed at the corners of a square of side a. The electric field intensity at the center of the square will be

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Solution

Step 1: Given data
A square of side a has charge of $+ q$ at the corners.
Step 2: Analyzing the diagram
In a square, the distance of each corner from the center will be equal.
The direction of charge from the point $A$ is represented as $e_{A}$ .
The direction of charge from the point $C$ is represented as $e_{C}$ .
The electric field due to point charge $q$ at a distance $r$ is given by Coulomb's law as $E = \frac{k q}{r^{2}}$
Demonstrating with the help of a diagram.
Step 3: Calculating electric field intensity at each corner
Electric field due to charge at $C$ = Electric field due to charge at $A$ i.e., $|\vec{E_{C}}| = |\vec{E_{A}}|$
But, there is a change in direction. Hence, $\vec{E_{C}} = - \vec{E_{A}}$
Similarly, Electric field due to charge at $D$ = Electric field due to charge at $B$ i.e., $\vec{E_{D}} = - \vec{E_{B}}$
Step 4: Calculating electric field intensity at the center of the square
${\vec{E}}_{c e n t e r} = {\vec{E}}_{A} + {\vec{E}}_{B} + {\vec{E}}_{C} + {\vec{E}}_{D} {\vec{E}}_{c e n t e r} = {\vec{E}}_{A} + {\vec{E}}_{B} - {\vec{E}}_{A} - {\vec{E}}_{B} [A s {\vec{E}}_{C} = - {\vec{E}}_{A}, {\vec{E}}_{D} = - {\vec{E}}_{B}] {\vec{E}}_{c e n t e r} = 0$
Hence, the electric field intensity at the center of the square will be $0$

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