Question

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force of the separation is increased to 25 cm?

Answer 1

The force exerted by the charged particles at a distance of 25 cm on each other is 12.8 N

Explanation:

Step 1:

Given  

F = 20 N.

r₁ = 20 cm. = 0.2 m

F is Force exerted between the charges

r₁ is Distance between the Charges

Let m1 and m2 be the masses of the two charges.

Step 2:

Now, Using the Formula,

$\begin{equation}\begin{aligned}&F=\frac{\left(\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}\right)}{\left(\mathrm{r}_{1}\right)^{2}}\\&F\left(r_{1}\right)^{2}=\left(G m_{1} m_{2}\right)\\&20 \times 0.2^{2}=\left(\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}\right)\\&20 \times 0.04=\left(\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}\right)\\&0.8=\left(\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}\right)\end{aligned}$

G m₁ m₂ = 0.8……………………….eqn(1)

Step 3:

Now, if the load separation increases to 25 cm (or 0.25 m) then,

$\begin{equation}F=\frac{\left(\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}\right)}{0.25^{2}}$

From eqn (1) G m₁ m₂ = 0.8

$\begin{equation}\begin{aligned}&F=\frac{0.8}{0.25^{2}}\\&F=\frac{0.8}{0.0625}\\&F=12.8 \mathrm{N}\end{aligned}$

Therefore, the force exerted by the charged particles at a distance of 25 cm on each other is 12.8 N

Answer 2

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12.8 N

will be the answer

hope it help