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Step-by-Step Solution
Step 1: Understand the Charge Distribution
The cube has side length $a$. At each vertex (corner) of the cube, there is a charge $+Q$, except at the origin where the charge is $-Q$. We want to find the electric field at the center of the cube.
Step 2: Concept of Replacement of Charges
We can think of the $-Q$ charge at the origin as being composed of two parts: $+Q$ and $-2Q$. In other words,
$-Q = +Q + (-2Q)\,.
This helps us imagine that there is effectively a $+Q$ charge at the origin (making the total of $+Q$ at every corner), plus an additional $-2Q$ at the origin. This is a useful mathematical approach for superposition of fields.
Step 3: Electric Field Contribution of $+Q$ at Every Corner
If we consider every corner (including the origin) as having $+Q$, then the distribution of identical charges at the corners of a symmetric cube leads to a net electric field of zero at the center. This is because the contributions from each $+Q$ charge cancel out perfectly due to symmetry.
Step 4: Net Electric Field at the Center
After recognizing the symmetry cancellation from the $+Q$ charges, the only remaining contribution to the electric field at the center is due to the extra $-2Q$ charge placed at the origin. Hence, we only need to calculate the field due to this $-2Q$ charge at the center of the cube.
Step 5: Determine the Position Vector of the Center
The center of the cube relative to the origin has coordinates
$(\tfrac{a}{2}, \tfrac{a}{2}, \tfrac{a}{2})$. Therefore, its position vector can be written as
$\vec{r} = \frac{a}{2}\,\bigl(\hat{x} + \hat{y} + \hat{z}\bigr)\,.
The magnitude of this position vector is
$\bigl|\vec{r}\bigr| = \sqrt{\Bigl(\frac{a}{2}\Bigr)^2 + \Bigl(\frac{a}{2}\Bigr)^2 + \Bigl(\frac{a}{2}\Bigr)^2}
= \sqrt{3 \left(\frac{a}{2}\right)^2}
= \frac{a\,\sqrt{3}}{2}\,.
Step 6: Apply the Formula for Electric Field
The electric field at a distance $r$ from a point charge $q$ is given by
$\vec{E} = \frac{1}{4\pi \varepsilon_0} \,\frac{q}{r^2} \,\hat{r}\,,
where $\hat{r}$ is the unit vector in the direction from the charge to the point of interest. For our $-2Q$ charge at the origin, $q = -2Q$, and $r = \frac{a\,\sqrt{3}}{2}$. The direction $\hat{r}$ is the unit vector along
$\bigl(\hat{x} + \hat{y} + \hat{z}\bigr)$.
Thus,
$\vec{E}
= \frac{1}{4\pi \varepsilon_0} \,\frac{-2Q}{\Bigl(\frac{a\sqrt{3}}{2}\Bigr)^2} \,\hat{r}
= \frac{1}{4\pi \varepsilon_0} \,\frac{-2Q}{\frac{3a^2}{4}} \,\hat{r}\,.
Step 7: Introduce the Direction Vector Explicitly
We also need to multiply by the correct unit vector. The unit vector in the direction of
$\bigl(\hat{x}+\hat{y}+\hat{z}\bigr)$ is
$\frac{\frac{a}{2}(\hat{x}+\hat{y}+\hat{z})}{\frac{a\sqrt{3}}{2}}
= \frac{\hat{x}+\hat{y}+\hat{z}}{\sqrt{3}}\,.
Putting it all together and simplifying eventually gives
$\vec{E}
= \frac{-2Q}{3\sqrt{3}\,\pi\,\varepsilon_0\,a^2}\,\bigl(\hat{x} + \hat{y} + \hat{z}\bigr)\,.
Final Answer
$\displaystyle \vec{E} = \frac{-2Q}{3\sqrt{3}\,\pi\,\varepsilon_0\,a^2}\,\bigl(\hat{x} + \hat{y} + \hat{z}\bigr)\,.
