Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Note: If a conducting uncharged material is brought in contact with a charged surface then the charges are shared uniformly between the two bodies.
Given: Charge on sphere 1, q1 = 6.5 × 10-7 C
Charge on sphere 2, q2 = 6.5 × 10-7 C
Charge on sphere 3, q3 = 0
Step 1: The uncharged sphere is brought in contact with sphere 1. Since sphere 1 has charge ‘q’, it gets distributed among sphere 1 and sphere 3.
Now, charge on sphere 1 = q/2
Charge on sphere 2 = q/2
At this point the sphere 3 which was initially uncharged has a charge “q/2”.
Step 2: Now sphere 3 is brought in contact with sphere 2 due to which 1/4 × q will flow from sphere 2 to sphere 3. Now sphere 2 and sphere 3 have “3/4 × q” charge.
Now, q1 = 1/2 × 6.5 × 10-7 C
⇒ 3.25 × 10-7 C
q2 = 3/4 × 6.5 × 10-7 C
⇒ 4.87 × 10-7 C
q3 = 3/4 × 6.5 × 10-7 C
⇒ 4.87 × 10-7 C
We know that,
Where, F= mutual force of attraction
q1 = charge on sphere 1
q2 = charge on sphere 2
r = distance between centres
Where, ε0 is the permittivity of the free space.
Plugging the values of q1, q2 and r in equation (1), we get
⇒ F = 5.703 × 10-3 N
The repulsive force between sphere 1 and sphere 2 is 5.703 × 10-3 N.
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Physics - Board Papers
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