Problem Analysis

We are tasked with finding the electrostatic potential at the vertex $C$ of a uniformly charged right-angled triangular lamina $ABC$. The right angle is at $C$. The acute angle at vertex $A$ is $\theta$. The potentials at vertices $A$ and $B$ are given as $V_A$ and $V_B$, respectively.

This problem can be solved elegantly using dimensional analysis (scaling arguments) and the principle of superposition. The electric potential $V$ at a specific geometric point (like a vertex) of a uniformly charged plane figure of characteristic linear dimension $L$ and uniform surface charge density $\sigma$ scales linearly with $L$.

Key Principle: For a shape of a given geometry, the potential at a homologous point is proportional to its linear size. Thus, $V \propto L$.

Step-by-Step Derivation

Let the hypotenuse $AB$ have length $c$. Since the triangle is right-angled at $C$ and $\angle A = \theta$, the side lengths are:

• Hypotenuse $AB = c$
• Side $AC = b = c \cos\theta$
• Side $BC = a = c \sin\theta$

Let $V(\alpha)$ be the potential at a vertex with angle $\alpha$ for a “standard” right-angled triangle with hypotenuse of unit length. Since potential scales linearly with size ($V \propto L$), the potentials for our triangle of size $c$ are:

• At vertex $A$ (angle $\theta$): $V_A = c \cdot V(\theta)$
• At vertex $B$ (angle $90^\circ – \theta$): $V_B = c \cdot V(90^\circ – \theta)$
• At vertex $C$ (angle $90^\circ$): $V_C = c \cdot V(90^\circ)$

Decomposition Method

We can divide the original triangle $ABC$ into two smaller right-angled triangles by dropping a perpendicular from vertex $C$ to the hypotenuse $AB$ at point $D$.

This creates two new triangles: $\Delta ADC$ and $\Delta BDC$.

1. Triangle ADC:

• This triangle is similar to the original triangle $ABC$.
• Its hypotenuse is $AC = b = c \cos\theta$.
• The vertex $C$ in the original triangle is now split. In $\Delta ADC$, the vertex at $C$ has the angle $90^\circ – \theta$.
• This vertex ($C$ in $\Delta ADC$) is homologous to vertex $B$ of the original triangle.
• Therefore, the contribution to the potential at $C$ from the charge on $\Delta ADC$ is: $$ V_{C,1} = (\text{Scaling Factor}) \times (\text{Potential at equivalent vertex } B \text{ of original}) $$ $$ V_{C,1} = \frac{AC}{AB} V_B = \cos\theta \cdot V_B $$

2. Triangle BDC:

• This triangle is also similar to the original triangle $ABC$.
• Its hypotenuse is $BC = a = c \sin\theta$.
• In $\Delta BDC$, the vertex at $C$ has the angle $\theta$.
• This vertex ($C$ in $\Delta BDC$) is homologous to vertex $A$ of the original triangle.
• Therefore, the contribution to the potential at $C$ from the charge on $\Delta BDC$ is: $$ V_{C,2} = (\text{Scaling Factor}) \times (\text{Potential at equivalent vertex } A \text{ of original}) $$ $$ V_{C,2} = \frac{BC}{AB} V_A = \sin\theta \cdot V_A $$

Superposition

By the principle of superposition, the total potential at vertex $C$ is the sum of the potentials contributed by the two parts of the lamina ($\Delta ADC$ and $\Delta BDC$):

$$ V_C = V_{C,1} + V_{C,2} $$ $$ V_C = V_B \cos\theta + V_A \sin\theta $$