To store a total of 0.040 J of energy in the two identical capacitors shown, each should have a capacitance of:
A) 0.10 $\mu$F
B) 0.50 $\mu$F0.10 $\mu$F
C) 1.0 $\mu$F
D) 1.5 $\mu$F
E) 2.0 $\mu$F

Question

To store a total of 0.040 J of energy in the two identical capacitors shown, each should have a capacitance of: 
A) 0.10 $\mu$F 
B) 0.50 $\mu$F0.10 $\mu$F 
C) 1.0 $\mu$F 
D) 1.5 $\mu$F 
E) 2.0 $\mu$F

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the formula:
E = (1/2)CV^2
where E is the energy, C is the capacitance, and V is the voltage across the capacitor.

We are given that the total energy stored is 0.040 J. The capacitors are identical, so they have the same capacitance (let's call it C), and they are connected in series, so the voltage across each capacitor is half of the total voltage.

Let's find the total voltage first. Since the capacitors are in series, the total capacitance is:
1/Ctotal = 1/C + 1/C
1/Ctotal = 2/C
Ctotal = C/2
So the total capacitance is half of the capacitance of each capacitor.

Now, we can use the formula for energy to solve for the capacitance:
E = (1/2)CV^2
0.040 = (1/2)Ctotal(Vtotal/2)^2
0.040 = (1/2)(C/2)(Vtotal^2/4)
0.160 = (1/4)CVtotal(Vtotal^2)
Vtotal^2 = (0.160)/(1/4CVtotal)
Vtotal^2 = 0.640CVtotal
Vtotal = sqrt(0.640CVtotal)

Since the capacitors are identical, the voltage across each capacitor is half of the total voltage:
V = Vtotal/2 = sqrt(0.640CVtotal)/2

Now we can solve for C:
0.040 = (1/2)C(V^2)
0.040 = (1/2)C((sqrt(0.640CVtotal)/2)^2)
0.040 = (1/2)C(0.640C/4)
0.040 = (1/2)C(0.160C)
0.160C^2 = 0.040*2
C^2 = 0.050
C = sqrt(0.050)
C = 0.22 μF
Since the capacitors are identical, each capacitor should have a capacitance of 0.22 μF. The closest answer choice is C) 1.0 μF.

To Store a Total of 0 μ\muμ F B) 0 μ\muμ F010 μ\muμ

To Store a Total of 0 $\mu$ F
B) 0 $\mu$ F010 $\mu$