Question

When two spheres having 2Q and -Q charge are placed at a certain distance the force acting between them is F. Now they are connected by a conducting wire and again separated from each other. How much force will act between them if the separation now is the same as before?

Answer 1

Answer:F/8

Explanation:

Initial F=kq3q/rsquare

Final f=kqq/4r

Here most importent you should know about sharing of charges

If charges are touched to each sharing of charges take place i. E

Charge from higher value move to lower value untill they equilised

Generally it's formulla is given by (Q1+Q2)/2 with sign +ve or -ve

Here Q1=2q and Q2=-q so (2q-q)/2

q/2 for each (Q1=Q2 after connection)

Applying to general formula

f=kqq/4r

So F/f=8

F/8=f

Hope you understood my level best!