Solution to Question 53
Physics Principle: Electrostatic Potential Profile and Energy Conservation.
System Setup:
\(+Q\) sphere centered at \(x=0\).
\(-Q\) sphere centered at \(x=-x_0\).
Bead \(+q\) travels from \(-\infty\).
Stopping region: Inside \(+Q\) and Outside \(-Q\).
Range: \( R-x_0 < x < R \).
1. Stopping Position \(x\):
At the stopping point, the initial kinetic energy is fully converted to potential energy.
$$ \frac{1}{2}m u^2 = q V(x) $$
The potential \(V(x)\) in the stopping region is the sum of the constant potential inside the positive sphere and the variable potential outside the negative sphere:
$$ V(x) = \frac{kQ}{R} – \frac{kQ}{x+x_0} $$
Energy Balance:
$$ \frac{m u^2}{2q} = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{R} – \frac{1}{x+x_0} \right) $$
Solving for \(x\):
$$ x = \frac{R}{1 – \frac{2\pi\epsilon_0 m u^2 R}{qQ}} – x_0 $$
2. Range of Initial Velocity:
The bead must stop before it exits the positive sphere at \(x=R\).
Potential at \(x=R\): \( V(R) = \frac{kQ}{R} – \frac{kQ}{R+x_0} \).
Condition: \( \frac{1}{2} m u^2 < q V(R) \).
$$ \frac{1}{2} m u^2 < \frac{qQ}{4\pi\epsilon_0} \left( \frac{1}{R} - \frac{1}{R+x_0} \right) $$
$$ u < \sqrt{ \frac{qQ}{2\pi\epsilon_0 m} \left( \frac{x_0}{R(R+x_0)} \right) } $$
Velocity Range: \( u < \sqrt{ \frac{qQ x_0}{2\pi\epsilon_0 m R (R+x_0)} } \)