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?

Question

David Clayton · Accepted Answer

Initial charge on the C1:&amp;nbsp;qi1=C1Vi=100⋅50=5000pC&amp;nbsp;final charge:&amp;nbsp;qf1=C1Vf=100⋅35=3500pC&amp;nbsp;After connecting the capacitors, the charge on the second capacitor will be&amp;nbsp;q2:&amp;nbsp;The electric potentials are equal because the capacitors are connected in parallel:V2=35V&amp;nbsp;the capacitance of the second capacitor is:&amp;nbsp;C2=q2V2=150035=43pC

kaluitagf · Accepted Answer

The charge initially on the charged capacitor is given by q=C1,V0, where C1=100pF is the capacitance and V0=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 q1=C1V, where V=35V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q2=q−q1, where C2 is the capacitance of the second capacitor. Substituting C1V0 for q and C1V for q1, we obtain q2=C1(V0−V). The potential difference across the second capacitor is also V, so the capacitance is:  C2=q2V=V0−VVC1=50V−35V35V(100pF)=43pF

karton · Accepted Answer

Consider the initial charge on the capacitor&amp;nbsp;(q)=C1V0=(100PF)(50V)=5000pCThe voltage on the first capacitor is now&amp;nbsp;q1=C1V&amp;nbsp;when the battery is detached and the second capacitor wire is parallel to the first.&amp;nbsp;The potential difference across the first capacitor drops to&amp;nbsp;(V)=35V,The&amp;nbsp;nq1=(100pF)(35V)=3500pCSince the charge is conserved in this process, then thecharge on the second capacitor is&amp;nbsp;q2=q−q1=5000pC−3500pC=1500pC. The capcacitance of the second capacitor is&amp;nbsp;C2=q2/V=1500pC/35V=42.85pF≈43pF

A 100 pF capacitor is charged to a potential difference of 50 V, and t

Answered question

Answer & Explanation

New Questions in Other