Question

We have a pair of charged objects separated by 2 meters and other pair of objects separated by 3 meters. Which pair will suffer higher forces?

a. First pair
b. Second pair
c. Both suffer the same forces​

Answer 1

Answer:

Electric Force between charged objects is given by the formula:

$\implies E_f = \dfrac{K.Q.q}{r^2}$

Here,

Q, q refer to charges or charge carrying objects and 'r' refers to the distance between the bodies or charges.

According to the formula, we see that, 'r' is inversely proportional to the electric force generated between them. Hence,

As we increase 'r', the electric force decreases and the vice versa.

Now it is given that in the first pair, the separation between the charges is 2 meters and in the second pair it is 3 meters.

Using the inverse proportion idea, we get:

Force on First Pair > Force on Second Pair

Hence the First pair with separation 2 m will suffer higher electric forces.

Answer 2

Option-A

Explanation:

We know that:

$\pink{\sf{F_e = k_e\dfrac{q_1q_2}{r^2}}}$

In this formula,

• $\sf{F_e}$ = Force
• $\sf{k_e}$ = Coulomb's constant
• $\sf{q_1,q_2}$ = Magnitudes of charge
• $\sf{r}$ = Distance between the charges.

In this formula, we can observe that, electric force is inversely proportional to it's distance.

From this, two points are clear:

1. If the distance increases, the electric force decreases.
2. If the distance decreases, electric force increases.

Now, let's come to the question...

We have a pair of charged objects separated by 2 meters and other pair of objects separated by 3 meters.

While comparing 2 metres and 3 metres, 2 metres is the distance which had been decreased:

→ 2m < 3m

Since, 2m distance is lower than 3m distance, we can say that electric force is higher in the first pair than the second pair.

Hence, Option-A is the correct option.

First pair experience higher force.