Wednesday, 1 June 2016

WORKSHEET #7 CAPACITANCE



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