Two charges of equal magnitude are separated by some distance. If the charges are increased by 10%, to get the same force between them, their separation must be

The correct answer is option 2):(increased by 10% )

Concept:

According to Coulomb's law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

\(F = K\frac{{{q_1}\; × \;{q_2}}}{{{r^2}}}\)

where

F is the force between the charges

q₁ q₂  are the charges

r is the distance between them

K is the constant

Calculation:

Case 1:

\(F = K\frac{{{q_1}\; × \;{q_2}}}{{{r^2}}}\) -------(1)

Case 2:

charges are increased by 10%

F is the same force

r₁ is the distance between them

F= K \( 1.1× q_1 × q_2 × 1.1 \over r_1^2\)  ----------(2)

To get the relationship between r and r₁

equate 1 and 2

\(1 \over r^2\) = \(1.21 \over r_1^2\)

r₁ = 1.1×  r

so the distance increased by 10 %