ELECTROSTATICS cYU 1

Problem 1: Time for Charge Separation

Step-by-Step Solution

The equation of motion for a particle of mass $m$ and charge $q$ repelled by an identical particle is:

$$ m \frac{d^2r}{dt^2} = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} $$

We can rearrange this to group the physical constants:

$$ \frac{d^2r}{dt^2} = \left( \frac{q^2}{4\pi\epsilon_0 m} \right) \frac{1}{r^2} $$

Let $R$ be the characteristic length scale (initial separation) and $T$ be the characteristic time scale. Dimensionally, $[r] \sim R$ and $[t] \sim T$. Substituting these into the differential equation:

$$ \frac{R}{T^2} \propto \left( \frac{q^2}{m} \right) \frac{1}{R^2} $$ $$ T^2 \propto \frac{m R^3}{q^2} \implies T \propto \sqrt{\frac{m R^3}{q^2}} $$

We can now compare the two cases using this proportionality.

Case 1 (Original):

• Charges: $q \times q \rightarrow q^2$
• Initial separation: $r_0$
• Time taken: $t_0$

$$ t_0 = k \sqrt{\frac{m r_0^3}{q^2}} $$

Case 2 (Modified):

• Charges: $aq \times bq \rightarrow abq^2$
• Initial separation: $\eta r_0$
• Time taken: $t’$

$$ t’ = k \sqrt{\frac{m (\eta r_0)^3}{ab q^2}} $$

Taking the ratio $t’/t_0$:

$$ \frac{t’}{t_0} = \frac{\sqrt{\frac{m \eta^3 r_0^3}{ab q^2}}}{\sqrt{\frac{m r_0^3}{q^2}}} = \sqrt{\frac{\eta^3}{ab}} $$