The three charges $\frac{q}{2},q$ and $\frac{q}{2}$ are placed at the corners $A,D$ and $C$ of a square of side $a$ as shown in figure. The magnitude of electric field $(E)$ at the corner $B$ of the square, is ![JEE Main 2022 Physics, Electrostatics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a6e417cc3fb48d67cdb/question_1__q_648b5a6e417cc3fb48d67cdb__cdn-question-pool.getmarks.app__5ded6261-abb9-4c2e-bef2-043a34a5f104-D38__9012e23087_final_ppt_sync.png)

![JEE Main 2022 Physics, Electrostatics — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a6e417cc3fb48d67cdb/solution_1__q_648b5a6e417cc3fb48d67cdb__cdn-question-pool.getmarks.app__6c80369c-addb-4703-9cb3-91b69723e7c9-image__e5a6901b01_final_ppt_sync.png) ${E}_{A}=\frac{k(\frac{q}{2})}{{a}^{2}}=\frac{kq}{2{a}^{2}}$, ${E}_{C}=\frac{k(\frac{q}{2})}{{a}^{2}}=\frac{kq}{2{a}^{2}}$ and ${E}_{D}=\frac{k(q)}{{(\sqrt{2}a)}^{2}}=\frac{kq}{2{a}^{2}}$ As we can see from diagram, ${E}_{C}&{E}_{A}$ are perpendicular. Therefore, net electric field will be, $=\frac{kq}{2{a}^{2}}+\sqrt{{(\frac{kq}{2{a}^{2}})}^{2}+{(\frac{kq}{2{a}^{2}})}^{2}}=\frac{kq}{2{a}^{2}}+\frac{kq}{\sqrt{2}{a}^{2}} =\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(\frac{1}{\sqrt{2}}+\frac{1}{2})$

JEE Main 2022 — Physics Electromagnetism

2022mcqeasy

The three charges $\frac{q}{2},q$ and $\frac{q}{2}$ are placed at the corners $A,D$ and $C$ of a square of side $a$ as shown in figure. The magnitude of electric field $(E)$ at the corner $B$ of the square, is

JEE Main 2022 Physics, Electrostatics — question figure

Official previous-year question

Held on 28 Jun 2022 · Verified 6 Jul 2026.

Options

A
$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(\frac{1}{\sqrt{2}}+\frac{1}{2})$
B
$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(1+\frac{1}{\sqrt{2}})$
C
$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(1-\frac{1}{\sqrt{2}})$
D
$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(\frac{1}{\sqrt{2}}-\frac{1}{2})$

Did you get this right?

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

${E}_{A}=\frac{k(\frac{q}{2})}{{a}^{2}}=\frac{kq}{2{a}^{2}}$ , ${E}_{C}=\frac{k(\frac{q}{2})}{{a}^{2}}=\frac{kq}{2{a}^{2}}$ and ${E}_{D}=\frac{k(q)}{{(\sqrt{2}a)}^{2}}=\frac{kq}{2{a}^{2}}$

As we can see from diagram, ${E}_{C}&{E}_{A}$ are perpendicular.

Therefore, net electric field will be,

$=\frac{kq}{2{a}^{2}}+\sqrt{{(\frac{kq}{2{a}^{2}})}^{2}+{(\frac{kq}{2{a}^{2}})}^{2}}=\frac{kq}{2{a}^{2}}+\frac{kq}{\sqrt{2}{a}^{2}} =\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(\frac{1}{\sqrt{2}}+\frac{1}{2})$

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Options

$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(\frac{1}{\sqrt{2}}+\frac{1}{2})$

$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(1+\frac{1}{\sqrt{2}})$

$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(1-\frac{1}{\sqrt{2}})$

$\frac{q}{4\pi {\epsilon }_{0}{a}^{2}}(\frac{1}{\sqrt{2}}-\frac{1}{2})$

Solution

JEE Main 2022 Physics, Electrostatics — solution figure

${E}_{A}=\frac{k(\frac{q}{2})}{{a}^{2}}=\frac{kq}{2{a}^{2}}$ , ${E}_{C}=\frac{k(\frac{q}{2})}{{a}^{2}}=\frac{kq}{2{a}^{2}}$ and ${E}_{D}=\frac{k(q)}{{(\sqrt{2}a)}^{2}}=\frac{kq}{2{a}^{2}}$

As we can see from diagram, ${E}_{C}&{E}_{A}$ are perpendicular.

Therefore, net electric field will be,

JEE Main 2022 — Physics Electromagnetism

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Solution

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JEE Main 2022 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism