Respuesta :

cjmejiab

To solve this problem we will apply the concepts related to electric potential energy. This relationship shows us that energy is equivalent to half the product between the capacitance of the body by the squared voltage. Mathematically it can be expressed as,

[tex]E = \frac{1}{2} CV^2[/tex]

Here,

C = Capacitance

V = Voltage

E = Energy

The initial capacitance would then be given by

[tex]2mJ = \frac{1}{2} C (1.5)^2[/tex]

[tex]C = \frac{4*10^{-3}}{(1.5)^2}[/tex]

[tex]C = 1.78*10^{-3}F[/tex]

Energy stored in the capacitor with 3V is

[tex]E = \frac{1}{2} (1.78*10^{-3})(3)^2[/tex]

[tex]E = 8mJ[/tex]

Therefore the energy stored in the capacitor if it is charged to 3V is 8mJ

abidemiokin

The energy stored in the capacitor if it is charged to 3.0 V is 0.00801 Joules

The formula for calculating the energy stored in a capacitor is expressed as:

[tex]E=\frac{1}{2}CV^2[/tex]

If a capacitor charged to 1.5 V stores 2.0 mJ of energy, then;

[tex]2 \times 10^{-3}=\frac{1}{2}\times 1.5^2C\\0.002 = 1.125C\\C = \frac{0.002}{1.125} \\C = 1.78 \times 10^{-3}F[/tex]

Next is to determine the energy stored in the capacitor if it is charged to 3.0V

[tex]E = \frac{1}{2} \times 0.00178 \times 3^2\\E=0.00801Joules[/tex]

Hence the energy stored in the capacitor if it is charged to 3.0 V is 0.00801 Joules

Learn more here: https://brainly.com/question/21851402

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