Question

64 identical balls made of conducting material each having a potential of 10 mV are joined to form a bigger ball. The potential of a bigger ball is?

Answer 1

Since the balls are made of conducting material, the potential will be distributed uniformly over the surface of each ball. When the balls are joined to form a bigger ball, they will form a parallel combination, and the total charge will remain the same.
Let the potential of the bigger ball be V.
Then, by the principle of conservation of charge, we have:
total charge of the 64 balls = charge of the bigger ball
q = CV = 64C(0.01V)
where C is the capacitance of each ball and 0.01V is the potential of each ball.
Also, capacitance of each ball, C = 4πε₀r where r is the radius of each ball and ε₀ is the permittivity of free space.
Since all the balls are identical, r is the same for each ball.
So, we have:
q = 64(4πε₀r)(0.01V)
And, the capacitance of the bigger ball, C' is given by:
C' = 4πε₀R
where R is the radius of the bigger ball.
Since the balls are joined to form a bigger ball, the radius of the bigger ball, R, is equal to the radius of one ball multiplied by the cube root of 64.
So, we have:
R = r∛64 = 4r
Therefore, the capacitance of the bigger ball is:
C' = 4πε₀(4r) = 16πε₀r
Now, equating the expressions for q and C', we get:
64C(0.01V) = 16πε₀rV
or, V = $\frac{(64C)(0.01)}{(16πε₀r)}$
Substituting the value of capacitance C, we get:
V = $\frac{(64(4πε₀r))(0.01)}{(16πε₀r)}$
Simplifying, we get:
V = 0.16 V
Therefore, the potential of the bigger ball is 0.16 V.
Answer. 0.16V