Two metal spheres of radii 0.01m and 0.02m are given a charge of 15mC and 45mC respectively. They are then connected by a wire.the final charge on first sphere is...

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The final charge on 1st sphere is :- Q1= 20 mc
The final charge on 2nd Sphere is :- Q2= 40 mc .........

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**Solution:**

To find the final charge on the first sphere after it is connected to the second sphere by a wire, we can use the principle of charge conservation. According to this principle, the total charge before and after the process remains the same.

Let's denote the charge on the first sphere as Q1 and the charge on the second sphere as Q2. Initially, Q1 = 15 mC and Q2 = 45 mC.

**Charge Distribution:**

When the spheres are connected by a wire, charges redistribute themselves until they reach a state of equilibrium. In this state, the potential on the surfaces of both spheres becomes equal.

The charge distribution depends on the ratio of the radii of the spheres. In this case, the first sphere has a smaller radius (0.01 m) compared to the second sphere (0.02 m).

**Charge Distribution Scenario:**

When the spheres are connected, electrons flow from the sphere with a higher potential (Q2) to the sphere with a lower potential (Q1). Here's a step-by-step explanation of the charge distribution scenario:

1. Initially, the first sphere (Q1) has a charge of 15 mC, and the second sphere (Q2) has a charge of 45 mC.
2. Electrons from the second sphere (Q2) start flowing towards the first sphere (Q1) because it has a smaller radius and therefore a higher potential.
3. The flow of electrons continues until the potential on both spheres becomes equal.
4. Let's assume that after the redistribution of charges, the first sphere (Q1) acquires a charge of q1, and the second sphere (Q2) acquires a charge of q2.
5. Since the total charge is conserved, we have q1 + q2 = Q1 + Q2.
Substituting the given values, we get q1 + q2 = 15 mC + 45 mC = 60 mC.
6. The final charge on the first sphere (Q1) is q1, which can be found by rearranging the equation as q1 = q2 - Q2 + Q1.
Substituting the given values, we have q1 = 60 mC - 45 mC + 15 mC = 30 mC.

Therefore, the final charge on the first sphere after being connected to the second sphere is 30 mC.

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