Question

Two insulated charged copper Spheres A and B of identical size have charges qA and -3qA respectively. When they are brought in contact with each other and sepereted,what are the new charges on them?

Answer 1

Correct Question

Two insulated charged copper spheres A and B of identical size have charges qA and -3qA respectively. Another sphere C having charge qC is brought in contact with sphere A then with sphere B and finally removed. What are the new charges on them?

Solution-

Two insulated charged copper spheres A and B of identical size have charges qA and -3qA respectively.

Another sphere C having charge qC is brought in contact with sphere A then with sphere B and finally removed.

We have to find the new charges on both the spheres.

When sphere A and C come in contact with each other then the charge distributed on both of them is half of the original one.

New charge on sphere C = (Charge on sphere A + Charge on sphere C)/2

= (qA + qC)/2

Now, new charge on sphere C is qA.

When sphere C and B come in contact with each other.

Charge on B = (New charge on C + Charge on B)

= [(qA + qC)/2 - 3qA]/2

= [(qA + qC - 6qA)/2]/2

= (qC - 5qA)/4

Charge on A = (qA + qC)/2

Answer 2

Correct Question

Two insulated charged copper spheres A and B of identical size have charges qA and -3qA respectively. Another sphere C having charge qC is brought in contact with sphere A then with sphere B and finally removed. What are the new charges on them?

Answer

Let the new charge on C be qC

When A and C copper spheres are in contact .

Resultant charge on C = (Charge on A + Charge on C)/2

⇒ Charge on C =  $\bold{\dfrac{qA+qC}{2}}$

When B and C are in contact ,

Resultant charge on B = (  Charge on B + New Charge on C ) / 2

⇒ Charge on B = $\bold{ \dfrac{( -3qA) + ( qA + qC )/2}{2}}$

⇒ Charge on B = $\bold{\dfrac{( qC - 5qA ) }{2}}$

Now , charge on A = Charge on C =  $\bold{\dfrac{( qA + qC )}{2}}$

Sine we put first A & C in contact .