Three capacitors of 2 uF, 3 uF, and 6 uF are connected in series across 120 V. The voltage across the 6 uF capacitor is: <svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 380 120" style="max-width:380px;background:#faf9f5"> <title>Three capacitors 2uF, 3uF, 6uF in series across 120V</title> <!-- Wire and capacitors --> <line x1="10" y1="60" x2="60" y2="60" stroke="#14140f" stroke-width="1.2"/> <!-- C1 = 2uF --> <line x1="60" y1="35" x2="60" y2="85" stroke="#c6613f" stroke-width="2.5"/> <line x1="68" y1="35" x2="68" y2="85" stroke="#c6613f" stroke-width="2.5"/> <line x1="68" y1="60" x2="140" y2="60" stroke="#14140f" stroke-width="1.2"/> <text x="64" y="100" text-anchor="middle" font-size="9" fill="#c6613f">2μF</text> <text x="64" y="25" text-anchor="middle" font-size="8" fill="#14140f">60V</text> <!-- C2 = 3uF --> <line x1="140" y1="35" x2="140" y2="85" stroke="#c6613f" stroke-width="2.5"/> <line x1="148" y1="35" x2="148" y2="85" stroke="#c6613f" stroke-width="2.5"/> <line x1="148" y1="60" x2="230" y2="60" stroke="#14140f" stroke-width="1.2"/> <text x="144" y="100" text-anchor="middle" font-size="9" fill="#c6613f">3μF</text> <text x="144" y="25" text-anchor="middle" font-size="8" fill="#14140f">40V</text> <!-- C3 = 6uF --> <line x1="230" y1="35" x2="230" y2="85" stroke="#c6613f" stroke-width="2.5"/> <line x1="238" y1="35" x2="238" y2="85" stroke="#c6613f" stroke-width="2.5"/> <line x1="238" y1="60" x2="310" y2="60" stroke="#14140f" stroke-width="1.2"/> <text x="234" y="100" text-anchor="middle" font-size="9" fill="#c6613f">6μF</text> <text x="234" y="25" text-anchor="middle" font-size="8" fill="#788c5d" font-weight="600">20V</text> <!-- Voltage source labels --> <text x="10" y="55" font-size="9" fill="#14140f">+</text> <text x="315" y="55" font-size="9" fill="#14140f">−</text> <text x="160" y="115" text-anchor="middle" font-size="9" fill="#14140f">120V total; Q = 120μC each</text> </svg>

A capacitor of capacitance C is charged to V and connected to an identical uncharged capacitor. The loss in energy is:

A capacitor C1 = 5 uF is charged to 200 V. It is connected in parallel with an uncharged C2 = 15 uF. What percentage of the original energy is lost?

An air capacitor has C = 10 pF. Half the space is filled with a dielectric K = 7 (covering full area, half thickness). The new capacitance is:

A parallel plate capacitor (C = 10 uF, V = 100 V) is disconnected from the battery. The plates are pulled apart to double the separation. The new energy is: