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..Il vodafone TR 22:24 session. masteringphysics.com C Problem 244 A 2.90 pF cap

ID: 776479 • Letter: #

Question

..Il vodafone TR 22:24 session. masteringphysics.com C Problem 244 A 2.90 pF capacitor is charged to 500 v and a 3.55 uF capacitor is charged to 505 v Part A These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor? Enter your answers numerically separated by a comma. Part B What will be the charge on each capacitor? Enter your answers numerically separated by a comma Request Answer Part C What is the voltage for each capacitor if plates of opposite sign are connected? Express your answers using two significant figures. Enter your answers numerically separated by a comma Part D What is the charge on each capacitor if plates of opposite sign are connected? Express your answers using two significant figures. Enter your answers numerically separated by a comma

Explanation / Answer

The charge (Q) on each capacitor is the product of its capacitance (C) and the

voltage (V) across it: Q = C times V

So at first the 2.9 µF capacitor has a charge of 2.9 times 500, or 1450 micro

coulombs (µQ). The 3.55 µF capacitor has a charge of 3.55 times 505, or 1792.75 µQ.

When you connect them in parallel, the total charge stays the same, but the total

capacitance is the sum of their individual capacitances, or 2.9+3.55 = 6.45 µF and the total

charge is 1450 + 1792.75 or 3242.75 µQ. Some electrons flow from the 3.55 µF

capacitor (with higher voltage) to the 2.9 µF capacitor till the voltages are equal.

The final voltage (Vf) can be found by dividing the total charge by the

total capacitance, which is 3242.75 / 6.45 = 502.75 volts.

So the final voltage across both capacitors is 502.75 volts.

The charge on the 2.9 µF capacitor is (2.9)(502.75) = 1457.98 µQ

and the charge on the 6.8 µF capacitor is (3.55)(502.75) = 1784.77 µQ.