# Chapter 19 - Electric Potential Energy and the Electric Potential - Problems - Page 539: 46

$6.1\times10^{-5}\,C$

#### Work Step by Step

Recall: $E=\frac{qV}{2}$ or $V=2E/q$ Therefore, for capacitor A, we have $V=\frac{2\times5.0\times10^{-5}\,J}{11\times10^{-6}\,C}=9.1\,V$ The same voltage is applied between the plates of capacitor B, i.e., $V_{B}=9.1\,V$. Capacitance of capacitor B, $C_{B}=6.7\,\mu F=6.7\times10^{-6}\,F$ Then, $q_{B}=C_{B}V_{B}=6.7\times10^{-6}\,F\times9.1\,V=6.1\times10^{-5}\,C$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.