Question

The force between two changed particles is F. Now if each of the charge is doubled and the distance between them is halved the new force will be

Answer 1

Answer:-

Your Answer Is $\tt\dfrac{16\ k\ q_1\ q_2}{r^2}$ or 16 F.

Explanation:-

Given:-

• Forces between two charges is F.

• Let the two charges be $\tt q_1\ And\ q_2.$

• Let the distance between the charges be r.

Concept Used:-

Coulomb's Law:-

It states that the force between two point charges is directly proportional to the magnitude of the charges and inversely proportional to the square of distance between them.

$\\ \tt\implies F\propto q_1\times q_2$

$\\ \tt\implies F\propto\dfrac{1}{r^2}.$

$\\ \tt\implies F\propto\dfrac{q_1\ q_1}{r^2}.$

$\\ \boxed{\bf\implies F = K\dfrac{q_1\ q_2}{r^2}.}$

Solution:-

When the charges are $\underline{\tt q_1\ and\ q_2.}$

$\\ \bf\mapsto F = k\dfrac{q_1\ q_2}{r^2}. --- eq(1)$

When the charges are doubled and distance between them is halved.

• New charges will be $\tt 2q_1\ and\ 2q_2.$
• New distance will be $\tt\dfrac{r}{2}.$

So the new force will be:-

$\\ \tt = k\dfrac{2q_1\times 2q_2}{\bigg(\frac{r}{2}\bigg)^2}.$

$\\ \tt = k\dfrac{4\ q_1\ q_2}{\frac{r^2}{4}}.$

$\\ \tt = 16 \ k\dfrac{q_1 q_2}{r^2}.$

$\large\boxed{\bf{ = 16 F.}}}\ \ \ \ \ \bigg(\rm From\ eq(1).\bigg)$

Answer 2

Given :-

The force between two changed particles is F.

To Find :-

New Force

Solution :-

We know that

$\sf F =K \dfrac{q_1q_2}{r^2}$

The new charge =

$\sf 2q_1 \;and\; 2q_2$

$\sf F = k\dfrac{2q_1 \times 2q_2}{\bigg(\dfrac{r}{2}\bigg)^2}$

$\sf F = k\dfrac{2q_1 \times 2q_2}{\dfrac{r^2}{2^2}}$

$\sf F = k\dfrac{2q_1 \times 2q_2}{\dfrac{r^2}{4}}$

$\sf F = k\dfrac{2q_1 \times 2q_2}{r^2}\times 4$

$\sf F = k\dfrac{4 q_1q_2}{r^2} \times4$

$\sf F = 16k\dfrac{q_1q_2}{4}$

$\sf\dfrac{16k\dfrac{q_1q_2}{4}}{k\dfrac{q_1q_2}{4}}$

$\sf 16F$