Answer to Question #185617 in Electric Circuits for Hope

Question

A capacitor has a capacitance of 33.68 pF. What potential difference would be required to store 69.82 pC?

Answer 1

To be given in question

Charge(Q)=69.82PC =$69.82\times10^{-12} C$

Capacity(C)=33.68PF =$33.68\times10^{-12}F$

To be asked in question

Potential difference (∆v)=?

We know that

$∆V=\frac{Q}{C}\rightarrow(1)$

eqution (2)put value

$∆V=\frac{69.82\times 10^{-12}}{33.68\times10^{-12}} V$

$∆V=2.07304V$