1. System Configuration and Geometry

We consider a cube of side length $l$. A coordinate system is chosen with the origin $O$ at one vertex $(0,0,0)$. The center of the cube, point $C$, is located at $(l/2, l/2, l/2)$.

Three finite line charges (filaments) of length $l$ are arranged on three mutually skew edges of the cube as shown in the figure:

• Filament 1 (Along y-axis): Connects $(0,0,0)$ to $(0,l,0)$. Orientation: $\hat{j}$.
• Filament 2 (Parallel to z-axis): Connects $(l,0,0)$ to $(l,0,l)$. Orientation: $\hat{k}$.
• Filament 3 (Parallel to x-axis): Connects $(0,l,l)$ to $(l,l,l)$. Orientation: $\hat{i}$.

2. Electric Field Vector Calculation

The electric field $\vec{E}$ at a distance $r$ from the center of a finite line charge of length $l$ (where the point lies on the perpendicular bisector) is given by: $$ E = \frac{\lambda}{4\pi\epsilon_0 r} (\sin 45^\circ + \sin 45^\circ) = \frac{\sqrt{2}\lambda}{2\pi\epsilon_0 l} $$ Let $E_0 = \frac{\sqrt{2}\lambda}{2\pi\epsilon_0 l}$. The field direction is radially outward from the wire. We calculate the direction vector for each wire at the center $C(l/2, l/2, l/2)$.

Filament 1 (on y-axis):

The wire is along the y-axis ($x=0, z=0$). The perpendicular vector from the wire to the center is in the x-z plane: $$ \vec{r}_1 = \left( \frac{l}{2} – 0 \right)\hat{i} + \left( \frac{l}{2} – 0 \right)\hat{k} = \frac{l}{2}(\hat{i} + \hat{k}) $$ The unit vector for the field direction is: $\hat{u}_1 = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k})$. $$ \vec{E}_1 = E_0 \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}) $$

Filament 2 (parallel to z-axis, at $x=l, y=0$):

The wire is at $x=l, y=0$. The perpendicular vector from the wire to the center is in the x-y plane: $$ \vec{r}_2 = \left( \frac{l}{2} – l \right)\hat{i} + \left( \frac{l}{2} – 0 \right)\hat{j} = -\frac{l}{2}\hat{i} + \frac{l}{2}\hat{j} $$ The unit vector is: $\hat{u}_2 = \frac{1}{\sqrt{2}}(-\hat{i} + \hat{j})$. $$ \vec{E}_2 = E_0 \frac{1}{\sqrt{2}}(-\hat{i} + \hat{j}) $$

Filament 3 (parallel to x-axis, at $y=l, z=l$):

The wire is at $y=l, z=l$. The perpendicular vector from the wire to the center is in the y-z plane: $$ \vec{r}_3 = \left( \frac{l}{2} – l \right)\hat{j} + \left( \frac{l}{2} – l \right)\hat{k} = -\frac{l}{2}\hat{j} – \frac{l}{2}\hat{k} $$ The unit vector is: $\hat{u}_3 = \frac{1}{\sqrt{2}}(-\hat{j} – \hat{k})$. $$ \vec{E}_3 = E_0 \frac{1}{\sqrt{2}}(-\hat{j} – \hat{k}) $$

3. Net Electric Field

We sum the electric field vectors from all three filaments: $$ \vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 $$ $$ \vec{E}_{net} = \frac{E_0}{\sqrt{2}} \left[ (\hat{i} + \hat{k}) + (-\hat{i} + \hat{j}) + (-\hat{j} – \hat{k}) \right] $$ Grouping the components: