Question

Two charges of $40\mu C$ and $-20\mu C$ are placed distance apart. They are touched and kept at the same distance. The ratio of the initial to the final force between them is:

• A. 8 : 1
• B. 4 : 1
• C. 1 : 8
• D. 1 : 1

Answer 1

Answer: (A)

8 : 1

Given,

$Q_1=40\mu C$

$Q_2=-20\mu C$

$d=$ distance between two charge body

From the Coulomb force, the initial force

$F=\frac{Q_1{Q}_2}{4\pi {\epsilon }_0{d}^2}$ . . . . . .(1)

when the charge particle are touched and kept at the same distance then the charge is same,

$Q=\frac{Q_1+Q_2}{2}=\frac{40-20}{2}=10\mu C$

The final force

$F^{\prime }=\frac{Q^2}{4\pi {\epsilon }_0{d}^2}$. . .. . .. . . . .(2)

Divide equation (1) by (2), we get

$\frac{F}{F^{\prime }}=\frac{Q_1{Q}_2}{Q^2}=\frac{8}{1}$

$F:F^{\prime }=8:1$

The correct option is A.