Solution 1: Equilibrium on an Inclined Plane
Analysis of Forces:
Particle B is on an inclined plane of inclination $\alpha$. It is subject to three main forces in the plane of the slope:
- Component of Gravity: $mg \sin\alpha$ acting down the line of greatest slope.
- Electrostatic Repulsion ($F_e$): Acting along the line AB, pushing B away from A.
- Friction ($f$): Opposes the tendency of motion.
For equilibrium, the net force must be zero. We can resolve the forces into two components:
- Along the line AB: The electrostatic force $F_e$ can balance the component of gravity along this line ($mg \sin\alpha \cos\beta$). Since the distance $AB$ can be adjusted (charge B is “placed”, implying we find the condition where it can stay), we assume the radial forces balance.
- Perpendicular to line AB: The electrostatic force has no component here. Friction must balance the component of gravity acting perpendicular to AB.
Step 2: Force Balance Equation
The component of gravity acting perpendicular to the line AB tends to rotate B around A. This component is:
$$ F_{\perp} = (mg \sin\alpha) \sin\beta $$To prevent slipping, the static friction $f$ must balance this force:
$$ f = mg \sin\alpha \sin\beta $$Step 3: Limiting Friction Condition
The maximum available friction is proportional to the normal reaction $N$. On an inclined plane, $N = mg \cos\alpha$.
$$ f_{max} = \mu N = \mu mg \cos\alpha $$For equilibrium to be possible, the required friction must not exceed the maximum limit:
$$ f \le f_{max} $$ $$ mg \sin\alpha \sin\beta \le \mu mg \cos\alpha $$Canceling $mg$ (assuming non-zero mass):
$$ \sin\alpha \sin\beta \le \mu \cos\alpha $$ $$ \sin\beta \le \mu \frac{\cos\alpha}{\sin\alpha} = \mu \cot\alpha $$Step 4: Maximum Angle
The maximum value of $\beta$ corresponds to the equality:
$$ \sin\beta_{max} = \mu \cot\alpha $$ $$ \beta_{max} = \sin^{-1}(\mu \cot\alpha) $$