Two similar very small conducting spheres having charges $40 C$ and $- 20 C$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

Question

Two similar very small conducting spheres having charges

40 C

and

- 20 C

are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

Toppr Admin · Accepted Answer

Initially force between conducting spheres $F_{1} = \frac{K q_{1} q_{2}}{r^{2}} = \frac{K (40) (- 20)}{r^{2}}$ (i)
After contact total charge $= 40 - 20 = 20 μ C$
Now charge will be equally distributed
so $q_{1} = q_{2} = \frac{20}{2} = 10 μ C$
Finally force between spheres $F_{2} = \frac{K (10) (10)}{r^{2}}$ (ii)
Ratio of forces= $\frac{F_{1}}{F_{2}} = \frac{40 \times 20}{10 \times 10} = 8$

Two similar very small conducting spheres having charges 40C and −20C are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:8:14:11:81:1

Initially force between conducting spheres F1=Kq1q2r2=K(40)(−20)r2(i)After contact total charge=40−20=20μCNow charge will be equally distributedso q1=q2=202=10μCFinally force between spheresF2=K(10)(10)r2(ii)Ratio of forces=F1F2=40×2010×10=8

Two similar very small conducting spheres having charges $40 C$ and $- 20 C$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:
$8 : 1$
$4 : 1$
$1 : 8$
$1 : 1$