Answer:
6.9 x 10^-7 J
3.5 x 10^-7 J
Explanation:
<u>Identify the unknown: </u>
The energies stored in the capacitor before and after the plates are pulled farther apart
<u>List the Knowns: </u>
Voltage of the battery: V = 50 V
Area of the plates: A = 0.25 x 0.25 = 0.0625 m^2
Original distance between the plates: d = 1 mm = 10^-3 m
New distance between the plates: d = 2 mm = 2 x 10^-3 m
Permittivity of free space: ∈o = 8.85 x 10^-12 C^2/Nm^2-
<u>Set Up the Problem:
</u>
Capacitance of a parallel-plate capacitor:
C=∈o*A/d
Energy stored in a capacitor:
U_c=(1/2)*V^2*C
<u>Solve the Problem:
</u>
<u>Before the plates are pulled farther apart: </u>
C = 8.85 x 10^-12 x 0.0625/10^-3 = 5•53 x 10^-10 F
U_c = (1/2) x (50)^2 x 5.53 x 10^-10= 6.9 x 10^-7 J
<u>After the plates are pulled farther apart: </u>
C = 8.85 x 10^-12 x 0.0625/2*10^-3 = 2•77 x 10^-10 F
U_c = (1/2) x (50)^2 x 2•77 x 10^-10 = 3.5 x 10^-7 J
The energy decrease because the capacitance decrease, so the stored charge decrease and transferred to the battery
