JEE Main 2022 · 28 Jun · Shift 1 · Physics Q39

← JEE Main 2022 · 28 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 39 of 90

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

Square ABCD of side a: q/2 at A (top left), q at B (bottom left), q/2 at C (bottom right); D (top right) is empty.

\dfrac q{4\pi\varepsilon_0a^2}\left(\dfrac1{\sqrt2}+\dfrac12\right)

\dfrac q{4\pi\varepsilon_0a^2}\left(1+\dfrac1{\sqrt2}\right)

\dfrac q{4\pi\varepsilon_0a^2}\left(1-\dfrac1{\sqrt2}\right)

\dfrac q{4\pi\varepsilon_0a^2}\left(\dfrac1{\sqrt2}-\dfrac12\right)

Answer and worked solution

Official answer: A

From A and C (each $\frac q2$ at distance $a$ , at right angles): $\sqrt2\cdot\dfrac{kq}{2a^2}=\dfrac{kq}{\sqrt2a^2}$ along the diagonal BD, away from B.

From B (charge $q$ at $\sqrt2a$ ): $\dfrac{kq}{2a^2}$ along the same direction.

Total. $\dfrac q{4\pi\varepsilon_0a^2}\left(\dfrac1{\sqrt2}+\dfrac12\right)$ .

(A)

Topics:Electric field Coulomb's law and superposition

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JEE Main 2022 · 28 Jun · Shift 1, Physics question 39