Consider the circuit shown in the figure, with C1 = 5.62 YF and C2 = 7.44 yF, 2.00 pF C, 6.00pF 4+ - 90.0V @ (a) Find the equivalent capacitance (in pF) of the system. (4.47 L Jv F (b) Find the charge (in pC) on each capacitor. 5.62 WF capacitor [251 Jv e ( (d) 6.00 pF capacitor [' Jv 7.44 UF capacitor [141.3 J v pC 2.00 yF capacitor [ ] v 4 261 uc 141.3 pc Find the potential difference (in V) on each capacitor. 5.62 pF capacitor 6.00 pF capacitor 7.44 yF capacitor 2.00 pF capacitor [476.44 ']J Vv ['43.5 }4 vV [ 70,65 X Q The potential difference across each of the capacitors in the circuit is given by AV = - Use this relationship and the charge found in part (b) to find the desired values. V 18.99 T X The potential difference across each of the capacitors in the circuit is given by AV = % Use this relationship and the charge found in part (b) to find the desired values. V Find the total energy (in mJ) stored by the group. (0.0181 | % Recall the equation for the energy in a capacitor: = %Ceqdvz. In this case, you will need to use the equivalent capacitance you found in part (a). ml