both the point charges double and the distance is increased to 60 mm ?

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UpStudy AI · Accepted Answer

We start with Coulomb’s law:
  
$$
F = k\,\frac{q_1q_2}{r^2}.
$$

Assume that in the original situation the charges are $$ q_1 $$ and $$ q_2 $$ and the separation is $$ r_0 $$. In the new situation the charges become $$ 2q_1 $$ and $$ 2q_2 $$, and the new separation is $$ r' $$.

The new force is

$$
F' = k\,\frac{(2q_1)(2q_2)}{{r'}^2} = k\,\frac{4q_1q_2}{{r'}^2}.
$$

To see how the force changes, we compare the new force to the original force:

$$
\frac{F'}{F} = \frac{k\,\frac{4q_1q_2}{{r'}^2}}{k\,\frac{q_1q_2}{r_0^2}} = 4\,\frac{r_0^2}{{r'}^2}.
$$

If the distance is increased to $$ 60\,\text{mm} $$ and we assume that initially the distance was $$ 30\,\text{mm} $$ (since this is a common comparison when a distance “doubles”), then $$ r_0 = 30\,\text{mm} $$ and $$ r' = 60\,\text{mm} $$. Plug these into the ratio:

$$
\frac{F'}{F} = 4\,\frac{(30)^2}{(60)^2} = 4\,\frac{900}{3600} = 4\,\frac{1}{4} = 1.
$$

Thus, even though each charge doubles (which would tend to increase the force by a factor of $$4$$), the distance doubling increases the denominator by a factor of $$4$$ (since distance is squared), so the two effects cancel each other out. Consequently, the net effect is that the force remains unchanged.

When both charges double and the distance between them increases to 60 mm (from 30 mm), the force between them remains the same.

both the point charges double and the distance is increased to 60 mm ?

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