JEE PYQ: Electrostatics Question 4
Question 4 - 2021 (24 Feb Shift 1)
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:
(1) $\frac{2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x} + \hat{y} + \hat{z})$
(2) $\frac{Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x} + \hat{y} + \hat{z})$
(3) $\frac{-2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x} + \hat{y} + \hat{z})$
(4) $\frac{-Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x} + \hat{y} + \hat{z})$
Show Answer
Answer: (3)
Solution
If all corners had $+Q$, the electric field at centre would be zero by symmetry. The given condition is equivalent to all $+Q$ plus a $-2Q$ at the origin. The field at centre due to $-2Q$ at corner: distance from corner to centre $= \frac{a\sqrt{3}}{2}$. Field $= \frac{2KQ}{3a^2/4} = \frac{8KQ}{3a^2}$, directed from centre toward origin $(-\hat{x} - \hat{y} - \hat{z})$ direction. In vector form: $\vec{E} = \frac{-2Q}{3\sqrt{3}\pi\varepsilon_0 a^2}(\hat{x} + \hat{y} + \hat{z})$.
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