Can someone please answer the following question, showing all work, steps and so
ID: 3893468 • Letter: C
Question
Can someone please answer the following question, showing all work, steps and solutions? I will award all points to the most complete and correct answer. Thanks alot!!!!
An air-gap parallel plate capacitor is connected to a battery with voltage V = 12 V. Each plate has an area of 2800 cm2 and the separation between the plates is 0.500 cm. While the capacitor remains connected to the battery, we insert a dielectric ( = 3.0) into the gap of the capacitor, filling one quarter of the volume as shown below.
A) What is the new capacitance?
B) How much energy is stored in the capacitor with the dielectric in place?
Explanation / Answer
capacitance with dieelctric upto 1/4 volume of the slab is C = (3/4+ k) eoA/d
C = (3/4+3)*8.85*10^-12 * 0.2800/0.005
C = 1.858 nC
b. eenergy U = 0.5 cv^2
U = 0.5 * 1.858*10^-9 * 12*12
U = 1.33*10^-7 Joules