Vikolers6

2022-01-06

A 100 pF capacitor is charged to a potential difference of 50 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to 35 V, what is the capacitance of this second capacitor?

David Clayton

Initial charge on the C1:
${q}_{i}1={C}_{1}{V}_{i}=100\cdot 50=5000pC$
final charge:
${q}_{f}1={C}_{1}{V}_{f}=100\cdot 35=3500pC$
After connecting the capacitors, the charge on the second capacitor will be ${q}_{2}$
The electric potentials are equal because the capacitors are connected in parallel:
${V}_{2}=35V$
the capacitance of the second capacitor is:
${C}_{2}=\frac{{q}_{2}}{{V}_{2}}=\frac{1500}{35}=43pC$

kaluitagf

The charge initially on the charged capacitor is given by $q={C}_{1},{V}_{0}$, where ${C}_{1}=100pF$ is the capacitance and ${V}_{0}=50V$ is the initial potential difference. After the battery is disconnected and the second capacitor wired in parallel to the first, the charge on the first capacitor is ${q}_{1}={C}_{1}V$, where $V=35V$ is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is ${q}_{2}=q-{q}_{1}$, where ${C}_{2}$ is the capacitance of the second capacitor. Substituting ${C}_{1}{V}_{0}$ for q and ${C}_{1}V$ for ${q}_{1}$, we obtain ${q}_{2}={C}_{1}\left({V}_{0}-V\right)$. The potential difference across the second capacitor is also V, so the capacitance is:
${C}_{2}=\frac{{q}_{2}}{V}=\frac{{V}_{0}-V}{V}{C}_{1}=\frac{50V-35V}{35V}\left(100pF\right)=43pF$

karton

Consider the initial charge on the capacitor $\left(q\right)={C}_{1}{V}_{0}=\left(100PF\right)\left(50V\right)=5000pC$
The voltage on the first capacitor is now ${q}_{1}={C}_{1}V$ when the battery is detached and the second capacitor wire is parallel to the first.
The potential difference across the first capacitor drops to $\left(V\right)=35V,$
The $n{q}_{1}=\left(100pF\right)\left(35V\right)=3500pC$
Since the charge is conserved in this process, then thecharge on the second capacitor is ${q}_{2}=q-{q}_{1}=5000pC-3500pC=1500pC$. The capcacitance of the second capacitor is ${C}_{2}={q}_{2}/V=1500pC/35V=42.85pF\approx 43pF$

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