# Problem: A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V.A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V.The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate.What is the final potential difference across the plates of the capacitors when they are connected in parallel?

###### FREE Expert Solution

The charge stored by a capacitor:

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

For capacitors in parallel, the equivalent capacitance is:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{n}}}}$

Charge on the 10.0 - μF capacitor is:

Q10 = CV = (10.0 × 10-6)(10.0) = 1.0 × 10-4 C

84% (331 ratings) ###### Problem Details

A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V.

A 5.0-µF capacitor is similarly charged so that the potential difference between its plates is 5.0 V.

The two charged capacitors are then connected to each other in parallel with positive plate connected to positive plate and negative plate connected to negative plate.

What is the final potential difference across the plates of the capacitors when they are connected in parallel?

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