Question

Two point charges 5 microcoulumb and 10 microcoulumb are separated by a distance r in air. If an additional charge of -4 microcoulumb is given to each by what factor does the force between the charges change

Answer 1

Answer:

let first the force is F then F=k×5×10^-6×10×10^-6/r^2 and after addition -4micro coulamb then charge is 5-4=1micro coulamb, 10-4=6 micro coulamb and new F is F'=k×1×10^-6×6×10^-6/r^2then divide both equation after dividation k ,r^2 and 10^-6 is cancelled then F/F'=5×10/1×6 F'=3F/25

Answer 2

Given : Two point charges 5μC and 10μC are seperated by a distance r in air. if an additional charge -4μC is given to each.

To find : by what factor does the force between the charge change.

solution : initial force between them, F = K(5μC)(10μC)/r² ........(1)

now - 4μC charge is given to each.

new charges are ; (5μC - 4μC) = 1μC

(10μC - 4μC) = 6μC

new force between them, F' = k(1μC)(6μC)/r² .......(2)

from equations (1) and (2) we get,

F/F' = (5μC)(10μC)/(1μC)(6μC) = 50/6 = 25/3

F' = 3/25 F

percentage change, ∆F = (F' - F)/F × 100

= (3/25 F - F)/F × 100

= -22/25 × 100

= -88 %

hence, force is reduced by 88 % of its initial value.