A parallel-plate capacitor has plates of area 0.12 m^2 and a separation of 1.2 c
ID: 1629712 • Letter: A
Question
A parallel-plate capacitor has plates of area 0.12 m^2 and a separation of 1.2 cm. A battery charges the plates to a potential difference of 120 V and is then disconnected. A dielectric slab of thickness 0.4 cm and dielectric constant 4.8 is then placed between the plates. a) Find the capacitance before the slab inserted. b) What is the capacitance with the slab in place? c) With the slab in place what is the potential difference across the plates? d) How much potential energy changes in the process of inserting the slab?Explanation / Answer
A) C1= Aeo/d, =0.12*8.85e-12 /0.012
= 8.85*10^-11 F
B) 1/C2 =
0.004/(4.8*0.12*8.85e-12) + 0.008/(0.12*8.85e-12) =8.318e9
C2 = 1.20*10^-10 F
C) V2 = Q/C2 = C1V1 /C2 = 8.85e-11*120/1.20e-10
= 88.5 V
D) change in PE = 0.5 C1V1^2 - 0.5C2V2^2
= 0.5* (8.85e-11*120^2 - 1.20e-10*88.5^2)
= 1.67*10^-7 J