Electrostatics

JEE Main level, single correct, medium

108 of 311

Question 108 of 311

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

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Answer: (C) −2Q33πϵ0a2(x^+y^+z^)\dfrac{-2Q}{3\sqrt{3}\pi {\epsilon }_{0}{a}^{2}}\left(\hat{x}+\hat{y}+\hat{z}\right)

We can replace −Q-Q charge at origin by +Q+Q and −2Q.-2Q.
Now due to +Q+Q charge at every corner of cube. Electric field at center of cube is zero so now net electric field at center is only due to −2Q-2Q charge at origin.
E⃗=kqr⃗r3=1(−2Q)a2(x^+y^+z^)4πϵ0(a23)3\vec{E}=\dfrac{kq\vec{r}}{{r}^{3}}=\dfrac{1\left(-2Q\right)\dfrac{a}{2}\left(\hat{x}+\hat{y}+\hat{z}\right)}{4\pi {\epsilon }_{0}{\left(\dfrac{a}{2}\sqrt{3}\right)}^{3}}
E⃗=−2Q(x^+y^+z^)33πa2ϵ0\vec{E}=\dfrac{-2Q\left(\hat{x}+\hat{y}+\hat{z}\right)}{3\sqrt{3}\pi {a}^{2}{\epsilon }_{0}}

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