Respuesta :

skyluke89
The charge on each plate of the capacitor is given by:
[tex]Q=C \Delta V[/tex]
where C is the capacitance of the capacitor, while [tex]\Delta V[/tex] is the potential difference applied on the two plates. Since we are using two batteries of 1.5 V, the total voltage applied is (assuming the two batteries are in series) 1.5 V+1.5 V= 3.0 V. Therefore, the charge on each plate is
[tex]Q=(1.0 F)(3.0 V)=3.0 C[/tex]
abidemiokin

If you charge a 1.0 f capacitor with two 1.5 volt batteries, the amount of charge on each of the plates is 3colombs

The formula for calculating the charge on a capacitor is expressed as:

[tex]Q = C \triangle v[/tex]

C is the capacitance of the capacitor in farads

Δv is the change in the potential difference

Given the following parameters

C = 1.0F

For a two 1.5V batteries, v = 2(1.5) = 3V

Substitute the given parameters into the formula as shown:

[tex]Q = 1.0 \times 2(1.5)\\Q=1 \times 3.0\\Q =3Coulumbs[/tex]

Hence the amount of charge on each of the plates is 3colombs.

Learn more here: https://brainly.com/question/15393720

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