Answer:
B. 8 J
Explanation:
From the question,
The voltage in each of the identical capacitor is given as
W = 1/2CV².................... Equation 1
Where Q = charge, V = potential difference, C = Capacitance of the capacitor.
make V the subject of the equation
V = √(2W/C)................ Equation 2
Given: W = 4 J,
Substitute into equation 2 and find the value of V in terms of C
V = √(8/C)
Since both capacitor were connected in parallel,
The total potential difference = potential difference of on of the capacitor = √(8/C)
Total capacitance = 2C.
using equation 1,
W = 1/2(√(8/C))²×2C
W = 8 J.
Hence the right option is B. 8 J
Hi there!
[tex]\large\boxed{\text{D. 2 J}}[/tex]
There is initially 4 J of energy in the system.
Recall the equation for the Electric Potential Energy of a capacitor:
[tex]U = \frac{1}{2}\frac{Q^2}{C}[/tex]
For capacitors in parallel, the charge ADDS UP. Since both capacitors are identical, each capacitor will receive HALF the charge.
Thus:
[tex]U' = \frac{1}{2}\frac{(\frac{1}{2}Q)^2}{C} = \frac{1}{8}\frac{Q^2}{C}[/tex]
This is ONE-FOURTH of the original potential energy (U = 4J), so:
[tex]U' = \frac{1}{4}(4J) = 1J[/tex]
Therefore, EACH capacitor has 1 J of energy, so, the total energy is the sum:
[tex]E_T = 1 J + 1 J + \boxed{\text{ D. 2 J}}[/tex]
