WORKSHEET # 7 CAPACITANCE
1.A parallel plate capacitor is
charged by a battery, which is then disconnected. A dielectric slab is then
inserted in the space between the plates. Explain what changes, if any, occur
in the values of:
(i) potential difference
between the plates
(ii)Electric field between the
plates, and
(iii)The
energy stored in the capacitor.
ANS:i)decreases ii)decreases iii)decreases
2. A parallel plate is charged
by a battery, When the battery remains connected. A dielectric slab is inserted
in the space between the plates. Explain what changes, if any, occur in the the
values of:
(i) Electric potential between
the plates
(ii) Electric field strength
between the plates
(iii ) Capacitance
(iv ) Charge on the the plate
(v ) Energy
stored in the capacitor? Justify your answer in each case.
ANS:
i)remains same ii)remains sameiii) increases iv) increase v)increases
3. A capacitor has some
dielectric between its plates and the capacitor is connected to a DC source.
The battery is now disconnected and then the dielectric is removed. State
whether the capacitance, the energy stored in it, electric field, charge stored
and the voltage will increase, decrease or remains constant.
ANS: Capacitance
decreases,Charge remains same,energy stored increases,V increases and E increases
4. A capacitor is connected across a battery.
(i) Why does each plate receive a charge of exactly the
same magnitude ?
(ii) Is this true even if the plates are of different
sizes?
ANS :i) Due to conservation of charge ii) yes
5. (i) How would you connect
two capacitors across a battery, in series or parallel, so that they store
greater (a) total charge and (b) total energy?
(ii) What is
the dielectric constant of a metal & why?
ANS:i) a)parallel
b)parallel ii) infinity
6. A
parallel plate capacitor is to be designed with a voltage rating 1KV using a
material of dielectric constant 3 and dielectric strength 107 Vm-1 .For safety we would like the field never to exceed, say 10%
of the dielectric strength. What minimum area of the plates is required to have
a capacitance of 50pF?
ANS:
18.8cm2
7. Two
Identical parallel plates ( air) Capacitors C1 and C2 have capacitance C each. The space
between their plates is now filled with dielectrics as shown. If the two
capacitors still have equal capacitance, obtain the relation between dielectric
constant K, K1, K2 .
ANS:K = ½
x(K1+K2)
8.The two plates of a parallel
plate capacitor are 4mm apart. A slab of dielectric constant 3 and thickness
3mm is introduced between the plates with its faces parallel to them. The
distance between the plates is so adjusted that the capacitance of the capacitor
become of its original value . What is the new distance between the plates ?
ANS: 8mm
9. A capacitor of unknown
capacitance is connected across a battery of V volt. The charge stored in it Is
360 μC. When potential across the is reduced by 120 V, The charge stored in it
becomes 120 μC. Calculate
(i) The potential V and the
unknown capacitance C.
(ii) What will be the charge
stored in capacitor, if the voltage applied had increased by 120 V.
ANS:i)180V and 2µF ii)600µC
10. The plates of a parallel
plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm.
The capacitor is charged by connecting it to a 400 V supply.
(i) How much electrostatic
energy is stored by the capacitor?
(ii) View this energy as stored
in the electrostatic field between the plates and obtain the energy per unit
volume u. Hence, arrive at the relation between u and the magnitude of electric
field E between the plates.
ANS: i) 2.55x 10 - 6 Jii) 2.25
X 10 - 5 m 3 and
½ εoE2
11. Show that the force on each
plate of a parallel plate capacitor has a magnitude equal to ½ QE, where Q is
the charge on the capacitor and E is the magnitude of electric field between
the plates. Explain the origin of factor 1/ 2 .
12. A network of four
capacitors each of 12 μF capacitance, if connected to a 500 V supply as shown
in figure . Determine
(i) Equivalent capacitance of
the network and
(ii) Charge
on each capacitor.
ANS:i)16µF ii) 2000µC and 6000
µC
13. Five capacitors of capacitance 10μF each are
connected with each other, as shown in fig. Calculate the total capacitance
between the points A and C.
ANS:10µF
14. Two metal plates form a
parallel plate capacitor. The distance between the plates is d. A metal sheet
of thickness d/2 and of the same area is introduced between the plates. What is
the ratio of the capacitance in the two cases?
ANS:2:1
15. Obtain
the equivalent capacitance of network in given figure. For a 300V supply,
determine the charge and voltage across each capacitor.
ANS: Total
capacitance = 66.7 pF and
Potential difference on C 1 = 100 V
Potential difference on C 2 = 50 V
Potential difference on C 3 = 50 V
Potential
difference on C 4 = 200 V