Answer:
0.256 J.
Explanation:
The formula for the energy stored in a capacitor is given as
E = 1/2CV².................... Equation 1
Where E = Energy stored in the capacitor, C = Capacitance of the capacitor, V = Voltage.
Make C the subject of the equation
C = 2E/V².................. Equation 2
Given: E = 64 mJ = 0.064 J, V = 1.5 V.
Substitute into equation 2
C = 2(0.064)/(1.5²)
C = 0.128/2.25
C = 0.0569 F.
If the capacitor where fully charged by 3.00 V.
E = 1/2CV²
Given: C = 0.0569 F, V = 3.0 V
Substitute into the equation above,
E = 1/2(0.0569)(3²)
E = 0.256 J.
256.05 mJ
The energy (E) stored in a capacitor when it is charged by a battery of voltage,V, is related to the capacitance, C, of the capacitor as follows;
E = [tex]\frac{1}{2}[/tex] x C x V² ---------------(i)
From the question;
When a 1.50V battery is used, the energy is 64.0mJ. This means that when;
V = 1.50V, E = 64.0mJ = 0.064 J
Substitute these values into equation (i) to find the capacitance of the capacitor as follows;
0.064 = [tex]\frac{1}{2}[/tex] x C x 1.5²
=> 0.064 = [tex]\frac{1}{2}[/tex] x C x 2.25
=> 0.064 = 1.125 x C
=> C = [tex]\frac{0.064}{1.125}[/tex]
Solve for C;
=> C = 0.0569F
Now, the we know the capacitance of the capacitor, let's calculate how much energy will be stored if the battery were 3.00V by substituting V = 3.00V and C = 0.0569 into equation (i) as follows;
E = [tex]\frac{1}{2}[/tex] x 0.0569 x 3.00²
E = [tex]\frac{1}{2}[/tex] x 0.0569 x 9.00
E = 0.25605 J
Multiply the result by 1000 to convert it to mJ. i.e;
E = (0.25605 x 1000) mJ
E = 256.05 mJ
Therefore, the amount of energy it would store if the capacitor were fully charged by a 3.00V battery is 256.05 mJ
