Dept. of Physics, NTNU
SIF4005 Fysikk
Autumn 2000
Tutoring: Wednesday 18. oct. 2000, 12.15-14.00, R1
Deadline for handing in: Friday 20 october, 12.00
Homework problem set 9
Exercise 1
A parallel plate capacitors consists of two rectangular plates
with sides a= 10cm and b = 50 cm located as illustrated in
Fig. 1.1. The separation between the plates, l, can be
adjusted, and is initially l=l1 3.0 mm. Initially, there is air
between the plates. The capacitor is charged to an electric
potential U1 = 300 V.
→
b
a
l
Figure 1.1 Illustration of parallel plate capacitor
a) Calculate the strength of the electrical field | E | between
the plates of the capacitor, and the capacitance C of the capacitor. What is the electrical field above/under the
capacitor (and why ?)?
The electrical connection to the emf. is disconnected after charging of the capacitor. The distance l is subsequently
increased to l = l2 = 6.0 mm to fit a plate of dielectric material of thickness 6 mm to be inserted between the plates.
The dielectric material fills the total space between the capacitor plates. The electrical potential is observed to be
reduced to 1/10 (10%) of the original potential.
b) Calculate the dielectric constant (relative permitivity) κ
of the material being inserted in the plate capacitor
The parallel plate capacitor (Fig. 1.1) is now being used to
determine the level for the liquid with dielectric constant κ
= 25 in a container. The separation between the plates are
now l = l1 = 3.0 mm, and the parallel plate capacitor is
located inside the container with the longest side in the
vertical direction, and the distance between the lower part
and the bottom of container being c = 3.0 cm (Fig. 1.2). The
level of liquid is going to be determined from
measurements of the capacitance to the capacitor, and we
need an equation relating the determined capacitance to the
liquid level.
air
b
x
liquid
c
c) Deduce an equation for the capacitance of the capacitor
as a function of the level of the liquid, x, determined
from the lower end of the capacitor. What is the liquid
level in the container when the capacitance has
increased tenfold ( 10 times) relative to the situation
without any liquid ?
Fig. 1.2. Illustration of parallel plate capacitor
for determination of level of liquid.
Exercise 2.
Three resistors are provided. They have equal resistance of 5 · 103 ohm (5 kΩ). Which equivalent
resistances can be made from these three resistors (you need to use all three in each combination)? Make
a graph of the combinations, and calculate the
equivalent resistance for each combination.
Calculate the currents in each resistor for each
combination when the arrangement is connected to
an ideal direct current source with emf of 12 V.
ε
C
Rekv
Which of equivalent resistor would you choose so
that the capacitor with capacitance C = 300 · 10-6 F
(300 µF) is discharged with a time-constant of 1
sec, when the switch B in the circuit (Graph at the
right) is opened following full charging of the
B
capacitor?