Two particles with charges `+3.72 mu C` and `+1.86 mu C` are some distance apart. If 20% of the charge is transferred from particle to second particle then the electrostatic force between them is

Options

A. decreases by 12%
B. increases by 12%
C. increases by 4%
D. decreases by 4%

Answer

B

Explanation

Given, charge on the first particle,
`Q_(1) = + 3.72 mu C` and charge on second particle,
`Q_(2) = 1.86 mu C`
Then the electrostatic force between charges,
`F_(1) = (kQ_(1)Q_(2))/(R^(2))`
`= (k)/(R^(2)) (3.72 xx 1.86) 10^(-12) N`
`F_(1) = (k)/(R^(2)) (6.9192 xx 10^(-12))`
If 20% of `Q_(1)` is given to `Q_(2)`,
`Q._(2) = Q_(2) + Q_(1) xx (20)/(100) = 1.86 + 3.72 xx 0.20`
`rArr Q._(2) = 2.604 mu C`
and `Q._(1) = Q_(1) xx (80)/(100) = 2.976 mu C`
Hence, `F_(2) = (k)/(R^(2))Q._(1).Q._(2) = (k)/(R^(2)) 2.976 xx 2.604 xx 10^(-12)`
`F_(2) = (k)/(R^(2)) 7.7495`.
So, % increment in `F_(2)`,
`(F_(2) - F_(1))/(F_(1)) xx 100 = (k)/(R^(2))((7.7495 - 6.9192) xx 10^(-12))/((k)/(R^(2)) xx 6.9192 xx 10^(-12)) xx 100`
`= 12%`
Hence, the correct option is (b).