Question

Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 16Fi. Find an expression for their new separation. (Use the following as necessary: qA, qB, ri, and ε0.)

Answer 1

Answer:

$r_{final}=r_{i}/4$

Explanation:

We know force between 2 charges is given by

$F=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r^{2}}$

It is given initial force is attractive of magnitude$F_{i}$

Let $r_{i}$ be the initial separation

Thus

$F_{i}=\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{i}^{2}}$

Now

Let $r_{f}$ be the final separation

Thus final force becomes

$\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{f}^{2}}$

It is given $F_{f}$ = $16F_{i}$...................(i)

Using values of  $F_{f}$ and $F_{i}$ in equation i we have

$16\times \frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{i}^{2}}$ = $\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{f}^{2}}$

thus

$thus\\\\(\frac{r_{i}}{r_{f}})^{2}=16\\\\\frac{r_{i}}{r_{f}}=4\\\\\therefore r_{f}=r_{i}/4$

Thus the final separation is one-fourth of the initial separation