Question

Q.22 Two point charges +10 µC and +20 µC repel each other with a force of 100 N. If a charge of –2 µC is added to each charge, then force between them will become Choose answer: 72 N 7.2 N 720 N 100 N

Answer 1

Given:
The repulsive force experienced between two charges of magnitude +10μC and +20μC is equal to 100N.

To Find:

The force between them when -2μC is added to each charge.

Solution:

Let the two charges be separated by a distance of $r$ meters from each other.

Using Coulomb's law, the force between the two charges is given by

   $F=k\frac{Qq}{r^2}$

⇒ $100=\frac{k}{r^2} *10*10^{-6}*20*10^{-6}$

⇒ $\frac{k}{r^2} =5*10^{11}N/C^2$

When -2μC is added to each charge, new charges will be 8μC and 18μC.

New force between them

   $F'=k\frac{Q'q'}{r^2}$

⇒ $F'=\frac{k}{r^2} *Q'q'$

⇒ $F'=5*10^{11}*8*10^{-6}*18*10^{-6}$

⇒ $F'=72N$

Hence, the force between the charges when -2μC is added to both the charges is equal to 72N.

Answer 2

Given: We have been given that the repulsive force between two point charges of magnitude +10 µC and +20 µC is 100 N.

To Find: Here, we have to find the force between these two point charges when -2 µC is added to each charge.

Solution: Let the distance between these two point charges be r meters.

According to Coulomb's law, the force between these two charges can be written as,

⇒$F = k\frac{Qq}{r^{2} } \\\\100 = \frac{k}{r^{2} } \times 10 \times 10^{-6} \times20 \times10^{-6} \\\\\frac{k}{r^{2} } =5\times10^{11} N/C^{2}$

Now, since -2 µC will be added to both the charges separately, the value of these charges will become +8 µC and +18 µC respectively.

Let the new force between these two charges be F'.

The value of F' will be,

$F' = k \frac{Q'q'}{r^{2} } \\\\F' = \frac{k}{r^{2} } \times Q'q'\\\\F'= 5 \times 10^{11} \times 8 \times 10^{-6} \times 18 \times 10^{-6} \\\\F' = 72 \ N$

Therefore, the new force after -2 µC is added to each of the two charges will be 72 N.

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