The energy stored on the plates will be 378.5 J. The combination of capacitors is in series.
A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.
An electric insulator or vacuum, such as glass, paper, air, or a semiconductor termed a dielectric, can be used as the non-conductive zone.
The complete question is;
"If this combination of capacitors is attached to a 9.00 V battery; how much energy will be 25 μF 15 μF stored on the plates?"
If the two capacitors are in series the equivalent capacitance will be;
[tex]\rm \frac {1}{C} = \frac{1}{C_1}+\frac{1}{C_2} \\\\ \rm \frac {1}{C} = \frac{1}{25}+\frac{1}{15} \\\\ C = 9.375[/tex]
Energy will be stored on the plates;
[tex]\rm E = \frac{1}{2}CV^2 \\\\ E= \frac{1}{2} \times 9.375 \mu F \times ( 9 \ V)^2\\\\E = 378.5 \ J[/tex]
Hence the energy stored on the plates will be 378.5 J.
To learn more about the capacitor, refer to the link: https://brainly.com/question/14048432.
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