Parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mmapart. The capacitor is connected to a 50.0-V battery. With the battery still connected, the plates are pulled apart to a separation of 2.00 mm. What are the energies stored inthe capacitor before and after the plates are pulled farther apart? Why does the energydecrease even though work is done in separating the plates.

6.9 x 10^-7 J  

3.5 x 10^-7 J

Explanation:

Identify the unknown:  

The energies stored in the capacitor before and after the plates are pulled farther apart  

List the Knowns:

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-

Set Up the Problem:  

Capacitance of a parallel-plate capacitor:  

C=∈o*A/d

Energy stored in a capacitor:  

U_c=(1/2)*V^2*C

Solve the Problem:  

Before the plates are pulled farther apart:  

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  

After the plates are pulled farther apart:  

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  

are there options? cause you ain’t. real specific

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.