eLECTROSTATICS BYU 4

Solution: Induced Charges on Pyramid Frame

Analysis:

This problem is solved using the principle of superposition. The final electric field points from corner A to corner C. This vector can be decomposed into two orthogonal components:

$$ \vec{E}_{AC} = \vec{E}_{AD} + \vec{E}_{AB} $$

We are given the induced charges for the field component along AD (let’s define this as the y-axis direction). Due to symmetry:

• For field along AD: $q_{DC} = q_1$, $q_{OC} = q_2$. By symmetry, $q_{AB} = -q_1$, $q_{OA} = -q_2$.
• For field along AB (rotated 90°): The roles of the rods shift. $q_{BC} = q_1$, $q_{OC} = q_2$ (relative to the new axis).

Since the resultant field is diagonal $\vec{E}_{AC} = \frac{1}{\sqrt{2}}\hat{E}_{AD} + \frac{1}{\sqrt{2}}\hat{E}_{AB}$, we scale and superpose the charges.