Day: March 8, 2019

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201903080359 Solution to 2005-AL-PHY-IIA-12

We assume that the charges on a small sphere can be treated as one point charge.

At first, P, Q, and R are of charge \pm q, \mp q, and 0.

Then, P and R are put in contact and share their charges evenly. They are now of charge \pm \displaystyle{\frac{q}{2}}. We keep them separated afterwards.

Later on, R and Q are put in contact and share their charges evenly. Each of them is hence of charge:

\displaystyle{\frac{\bigg( \pm \displaystyle{\frac{q}{2}}\bigg) +(\mp q)}{2}}=\mp\displaystyle{\frac{q}{4}}.

We keep them separated afterwards.

The initial magnitude  F of electrostatic force between P and Q is given by:

F=\bigg| \displaystyle{k\frac{(\pm q)(\mp q)}{r^2}} \bigg|=k \displaystyle{\frac{q^2}{r^2}}

The final magnitude F' of electrostatic force between them is related to the initial F by:

F'=\bigg| \displaystyle{k\frac{ ( \pm \frac{q}{2}) (\mp \frac{q}{4})}{r^2}} \bigg|=\displaystyle{\frac{1}{8}}\bigg( k \displaystyle{\frac{q^2}{r^2}} \bigg) = \displaystyle{\frac{1}{8}}F

\begin{aligned} \quad & \quad & \textrm{Initially} & \quad & \rightarrow & \quad & P\textrm{ and }R\textrm{ in touch} &\quad & \rightarrow & \quad & R\textrm{ and }Q\textrm{ in touch}\\ P & & \pm q & & & & \pm \displaystyle{\frac{q}{2}} & & & & \boxed{\pm \displaystyle{\frac{q}{2}}} \\ Q & & \mp q & & & & \boxed{\mp q} & & & & \mp \displaystyle{\frac{q}{4}}\\ R & & 0 & & & & \pm \displaystyle{\frac{q}{2}} & & & & \mp \displaystyle{\frac{q}{4}} \end{aligned}

And the answer is B.