A parallel-plate capacitor has plates of area 0.10 m 2 and a separation of 2.60
ID: 2132095 • Letter: A
Question
A parallel-plate capacitor has plates of area 0.10 m2 and a separation of 2.60 cm. A battery charges the plates to a potential difference of 100 V and is then disconnected. A dielectric slab of thickness 4 mm and dielectric constant 4.8 is then placed symmetrically between the plates.
(a) What is the capacitance before the slab is inserted?
1
(b) What is the capacitance with the slab in place?
2
(c) What is the free charge q before the slab is inserted?
3
(d) What is the free charge q after the slab is inserted?
4
(e) What is the magnitude of the electric field in the space between the plates and dielectric?
5
(f) What is the magnitude of the electric field in the dielectric itself?
6
(g) With the slab in place, what is the potential difference across the plates?
7
(h) How much external work is involved in the process of inserting the slab?
Explanation / Answer
a) capacitance = Co = e0A/d
= 8.854 x 10-12 x 0.10/ 0.026
= 3.41 x 10-11 F
b) C = e0Ak / [k(d-t)] + t
= 8.854 x 10-12 x 0.10 x 4.8 / [4.8(0.026 - 0.004)] +0.004
= 3.88 x 10-11 F
c) q = C0V = 3.41 x 10-9 C
d) q = same
e) E = q/e0A = 3851.37 V/m
f) E' = E/k = 802.37 V/m
h) W = q^2/2 [1/C0 - 1/C] = 2.07 x 10-8 J