Question

two small metal spheres a and b have different electric potentials (a has a higher potential). part a: describe mathematically what happens if you connect them with a wire. let the charges of the two spheres after connection be $q_a$ and $q_b$ and radiuses be $r_a$ and $r_b$. express your answer in terms of the variables $r_a$ and $r_b$. part b: describe in words what happens if you connect them with a wire. if the two spheres have the same radius, then negative charge must flow from a to b. if the two spheres have the same radius, then negative charge must flow from b to a.

Explanation:

Step1: Recall electric - potential formula

The electric potential of a charged spherical conductor is given by $V = \frac{kq}{r}$ (for points outside or on the surface). When two conductors are connected by a wire, charge will flow until their potentials are equal, i.e., $V_A'=V_B'$. So, $\frac{kq_A'}{R_A}=\frac{kq_B'}{R_B}$.

Step2: Solve for the ratio of charges

Cross - multiply the equation $\frac{kq_A'}{R_A}=\frac{kq_B'}{R_B}$ to get $\frac{q_A'}{q_B'}=\frac{R_A}{R_B}$.

For Part B:
When two conductors of the same radius ($R_A = R_B$) are connected by a wire, and initially $V_A>V_B$, since $V=\frac{kq}{r}$, a positive charge will flow from the sphere with higher potential (A) to the sphere with lower potential (B) until their potentials are equal. In terms of negative charge flow, negative charge will flow from B to A.

Answer:

Part A: $\frac{q_A'}{q_B'}=\frac{R_A}{R_B}$
Part B: If the two spheres have the same radius, then negative charge must flow from B to A.