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Multiple Choice
Metal sphere A is charged until it has a potential of . Metal sphere B is initially uncharged, and has a radius twice the radius of sphere A. A small wire is used to connect spheres A and B, so that they form one continuous equipotential surface. What is the final potential on the surface of sphere B?
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Verified step by step guidance
1
Understand that when two conductive spheres are connected by a wire, they will share charge until they reach the same electric potential. This is because the wire allows free movement of charge between the spheres.
Recall that the potential \( V \) of a sphere is given by \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius of the sphere.
Since sphere B has twice the radius of sphere A, its potential will be affected by this difference in radius once the spheres are connected.
Set up the equation for the final potential \( V_f \) of both spheres after they are connected: \( V_f = \frac{k(Q_A + Q_B)}{r_A + r_B} \). Here, \( Q_A \) is the initial charge on sphere A, and \( Q_B \) is initially zero.
Solve for \( V_f \) using the given initial potential of sphere A (200 V) and the relationship between the radii of the spheres. This will give you the final potential on the surface of sphere B.