Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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? - Physics

Sum

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?

#### Solution

Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other.
So, $F_1 = \frac{1}{4\pi \in_0} \cdot \frac{q^2}{r_1^2}$

Also, $F_2 = \frac{1}{4\pi \in_0} \cdot \frac{q^2}{r_2^2}$

According to the question, we have :

$\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}$

$= \frac{20 \times 20}{25 \times 25} = \frac{16}{25}$

$\therefore F_2 = \frac{16}{25} \times F_1$

$\Rightarrow F_2 = \frac{16}{25} \times 20$

$\Rightarrow F_2 = 12 . 8 N \approx 13 . 0 N$

Therefore, the two charged particles will exert a force of 13.0 N on each other, if the separation is increased to 25 cm.

Concept: Work Done by a Constant Force and a Variable Force
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 4 The Forces
Exercise | Q 8 | Page 63
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