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

The three charges q 2,q and q 2… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

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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,

Options

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Solution

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Options

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Solution

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Mechanics

Modern Physics

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Waves & Oscillations

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