The three charged particles in the figure below are at the vertices of an isosceles -[Free] B56
The three charged particles in the figure below are at the vertices of an isosceles triangle (where d=3.00 cmd = 3.00 \, \text{cm}d=3.00cm). Taking q=8.20 μCq = 8.20 \, \mu\text{C}q=8.20μC, calculate the electric potential at point A, the midpoint of the base.
![The three charged particles in the figure below are at the vertices of an isosceles -[Free] B56](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20640%20159'%3E%3C/svg%3E)
🔷 Electric Potential at the Midpoint of the Base
We calculate the electric potential at point A, the midpoint of the base of an isosceles triangle formed by three charges:
- One positive charge at the top vertex
- Two negative charges at the bottom corners
- Distance between the midpoint and each base corner: d = 3.00 cm = 0.0300 m
- Charge magnitude: q = 8.20 μC = 8.20 × 10⁻⁶ C
🌟 Step 1 — Concept
The electric potential at a point is the scalar sum of the potentials from all point charges:
where:
- k = 8.99 × 10⁹ Nm²/C² (Coulomb constant)
- q = charge
- r = distance from charge to the point
🌟 Step 2 — Contributions from Each Charge
🔷 Left Corner Charge (−q)
Located at a distance d from A:
Substitute:
🔷 Right Corner Charge (−q)
Symmetric to the left corner:
🔷 Top Vertex Charge (+q)
This charge is at a distance 2d = 0.0600 m from A:
Substitute:
🌟 Step 3 — Total Electric Potential at A
Sum of all three contributions:
Compute:
✨ Final Answer
This negative result shows that the two negative charges at the base dominate the electric potential at the midpoint.