Equivalent capacitance for parallel capacitors:

$\overline{){{\mathbf{C}}}_{{\mathbf{eq}}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{2}}}}$

The charge stored in a capacitor:

$\overline{){\mathbf{Q}}{\mathbf{=}}{\mathbf{C}}{\mathbf{V}}}$

Energy stored in a capacitor:

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{V}}}^{{\mathbf{2}}}}$

C_{eq} = C_{1} + C_{2} = 2.00 + 7.40 = 9.40 μF(10^{-6}F/μF) = 9.40 × 10^{-6} F

Suppose you have a 9.00-V battery, a 2.00-μF capacitor, and a 7.40-μF capacitor

Find the charge and energy stored if the capacitors are connected to the battery in parallel

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