The figure shows a circuit section of four air-filled capacitors that is connect

Question

The figure shows a circuit section of four air-filled capacitors that is connected to a larger circuit. The graph below the section shows the electric potentialV(x)as a function of positionxalong the lower part of the section, through capacitor 4. Similarly, the graph above the section shows the electric potentialV(x)as a function of positionxalong the upper part of the section, through capacitors 1, 2, and 3. Capacitor 3 has a capacitance of 0.49 ?F. What are the capacitances of(a)capacitor 1 and(b)capacitor 2?

Explanation / Answer

From that you know that Cap1 has 2V across it, Cap2 has 5V across it, and Cap3 has 5V across it (5+5+2=12). Just from intuition you know right away that Cap2 must equal Cap3 in capacitance because equal voltage is dropping across both. Cap1 we actually need to whip out an equation tho: C=Q/V Capacitance equals Charge divided by Voltage. Capacitors in series must all be given the same charge when a voltage is applied. If Cap3 is 0.69uF and its voltage is 5 then you have a charge of Q=C*V (0.69uF * 5V). That equals 3.45e-6. Now use the equation again except on Cap1; C=Q/V (3.45e-6/2=1.72uF). I'm assuming here you know that the mu (u) is the same as 10^-6 or 1e-6. If you wanted to check you could use the equation on Cap2 to unsure that our intuition was correct.

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