JEE Main 2021 · 24 Feb · Shift 1 · Physics Q15

← JEE Main 2021 · 24 Feb · Shift 1

PhysicsSingle correct+4 / -1Question 15 of 90

A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is −Q. The electric field at the centre of cube is :

A cube of side a with one corner at the origin of the x, y, z axes; that corner carries −Q and the other seven corners carry +Q.

\dfrac{-Q}{3\sqrt3\pi\varepsilon_0a^2}(\hat x+\hat y+\hat z)

\dfrac Q{3\sqrt3\pi\varepsilon_0a^2}(\hat x+\hat y+\hat z)

\dfrac{-2Q}{3\sqrt3\pi\varepsilon_0a^2}(\hat x+\hat y+\hat z)

\dfrac{2Q}{3\sqrt3\pi\varepsilon_0a^2}(\hat x+\hat y+\hat z)

Answer and worked solution

Official answer: C

Superposition. Eight +Q charges give zero field at the centre; the actual set equals that plus −2Q at the origin.

Field of −2Q at distance $\frac{\sqrt3a}2$ , pointing towards the origin: $\dfrac{2Q}{4\pi\varepsilon_0\cdot\frac34a^2}=\dfrac{2Q}{3\pi\varepsilon_0a^2}$ along $-\frac{\hat x+\hat y+\hat z}{\sqrt3}$ .

(C)

Topics:Electric field Coulomb's law and superposition

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JEE Main 2021 · 24 Feb · Shift 1, Physics question 15