Two sphers of radii r1 and r2 having equal charges afe joined together...
Given:
Two spheres of radii r1 and r2 having equal charges are joined together with a copper wire.
Let the potential of each sphere after they are separated be v.
To find:
The initial charge on both spheres.
Solution:
When two spheres are joined with a copper wire, they become equipotential. Therefore, the potential of each sphere will be the same, i.e., v.
Using the formula for potential:
V = kQ/r
where V is the potential, k is Coulomb's constant, Q is the charge on the sphere, and r is the radius of the sphere.
For sphere 1:
V = kQ1/r1
Q1 = Vr1/k
For sphere 2:
V = kQ2/r2
Q2 = Vr2/k
Since the spheres have equal charges, Q1 = Q2.
Therefore, Vr1/k = Vr2/k
r1/r2 = V/V
Since the radii are known, the ratio of charges can be found.
Q1/Q2 = (r1/r2)^2
Now, the problem can be solved by substituting the given values of r1, r2, and V.
Answer:
The initial charge on both spheres was Q1 = Q2 = Vr/k, where r is the radius of the spheres and k is Coulomb's constant.
Two sphers of radii r1 and r2 having equal charges afe joined together...
V=K(Q1 + Q2) ÷ (R1 + R2)
Q1= Q2
V = 2KQ ÷ (R1 + R2)
Q = V (R1 + R2) ÷ 2K
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