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  • A cube of side 'a' has point charge +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 : <img alt="" src="https://cdn.entrance360.com/media/uploads/2021/03/04/116800.jpg"

A cube of side 'a' has point charge +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 :  
Option: 1 \frac{Q}{3\sqrt{3}\pi \varepsilon _0a^{2}}\left ( \hat{x} +\hat{y}+\hat{z}\right )
 
Option: 2 \frac{2Q}{3\sqrt{3}\pi \varepsilon _0a^{2}}\left ( \hat{x} +\hat{y}+\hat{z}\right )
Option: 3 \frac{-Q}{3\sqrt{3}\pi \varepsilon _0a^{2}}\left ( \hat{x} +\hat{y}+\hat{z}\right )  
Option: 4 \frac{-2Q}{3\sqrt{3}\pi \varepsilon _0a^{2}}\left ( \hat{x} +\hat{y}+\hat{z}\right )

Answers (1)

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\\ \text{We can replace -Q}$ charge at origin by $+Q$ and $-2 \mathrm{Q}$. Now due to $+\mathrm{Q}$ 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 $-2 \mathrm{Q}$ charge at origin. \\ \\ $\overrightarrow{\mathrm{E}}=\frac{\mathrm{kq} \overrightarrow{\mathrm{r}}}{\mathrm{r}^{3}}=\frac{1(-2 \mathrm{Q}) \frac{\mathrm{a}}{2}(\hat{\mathrm{x}}+\hat{\mathrm{y}}+\hat{\mathrm{z}})}{4 \pi \varepsilon_{0}\left(\frac{\mathrm{a}}{2} \sqrt{3}\right)^{3}}$ \\ \\ \\ $\overrightarrow{\mathrm{E}}=\frac{-2 \mathrm{Q}(\hat{\mathrm{x}}+\hat{\mathrm{y}}+\hat{\mathrm{z}})}{3 \sqrt{3} \pi \mathrm{a}^{2} \varepsilon_{0}}$

\\ \text{Here the position vector,}\ \vec{r}=\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})\\ \\ \left | \vec{r} \right |=\sqrt{\left(\frac{a}{2} \right)^2+\left(\frac{a}{2} \right)^2+\left(\frac{a}{2} \right)^2}=\sqrt{\frac{3a^2}{4}}=\frac{\sqrt{3}a}{2}

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