A capacitor, C₁= 5.00 μF, is charged up to 8.00 V. It is then connected to a second uncharged capacitor C₂ = 2.50 μF. The charge on C₂ after the system has come to equilibrium is

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The total charge on the system is conserved, so the initial charge on C₁ must be equal to the sum of the charges on C₁ and C₂ after they are connected. That is: Q1 = Q1’ + Q2’ where Q1 is the initial charge on C₁, Q1’ is the charge on C₁ after it is connected to C₂, and Q2’ is the charge on C₂ after the connection. The initial voltage on C₁ is 8 V, so its initial charge is: Q1 = C1V1 = (5.00 μF)(8.00 V) = 40.0 μC When the capacitors are connected, charge flows from C₁ to C₂ until the voltages across the two capacitors are equal. Let Q be the charge that flows from C₁ to C₂. Then the final charges on the two capacitors are: Q1’ = Q1 - Q Q2’ = Q The final voltage across each capacitor is the same, so we can set up an equation using the final charges: C1V = C2V’ where V is the final voltage across both capacitors. Substituting in the expressions for the final charges and solving for Q, we get: Q = Q1 - Q1’ = C1V - C2V’ = (5.00 μF)(V) - (2.50 μF)(V) = 2.50 μF V Substituting the voltage V = Q/(C1 + C2) = Q/(5.00 μF + 2.50 μF) = Q/7.50 μF, we get: Q = (2.50 μF)(Q/7.50 μF) = 0.833Q So the charge on C₂ is Q2’ = Q = 0.833Q1’. Substituting Q1’ = Q1 - Q = (2/3)Q into this expression, we get: Q2’ = (5/6)Q1 = (5/6)(40.0 μC) = 33.3 μC Therefore, the answer is B) 13.3 μC, since that is the charge on C₂ after the connection.

A Capacitor, C1 = 5

A Capacitor, C₁= 5