Q11.Two identical thin metal plates has charge π1 and π2 respectively such that π1 > π2. The plates were brought close to each other to form a parallel plate capacitor of capacitance πΆ. The potential difference between them is : π1 + π2 π1 - π2 (1) (2) πΆ πΆ π1 - π2 2π1 - π2 (3) (4) 2πΆ πΆ
What This Question Tests
This question tests the understanding of charge distribution on parallel plates and how to calculate the potential difference for a capacitor formed by them, considering the net charge on the inner surfaces.
Concepts Tested
Formulas Used
V = Q/C
E = Ο/Ξ΅β
Ο = Q/A
π NCERT Sections This Tests
2.5 β A Parallel Plate Capacitor With Air Between The Plates Has A
Physics Class 11 Β· Chapter 2
2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10β12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
2.14 β Combination Of Capacitors
Physics Class 11 Β· Chapter 2
2.14 COMBINATION OF CAPACITORS We can combine several capacitors of capacitance C1, C2,β¦, Cn to obtain a system with some effective capacitance C. The effective capacitance depends on the way the individual capacitors are combined. Two simple possibilities are discussed below. 2.14.1 Capacitors in series Figure 2.26 shows capacitors C1 and C2 FIGURE 2.26 Combination of two combined in series. capacitors in series. The left plate of C1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and βQ , respectively. It then follows that the right plate of C1 has charge βQ and the left plate of C2 has charge Q. If this was not so, the net charge on each capacitor would not be zero. This would result in an electric field in the conductor connecting C1and C2. Charge would flow until the net charge on both C1 and C2 is zero and there is no electric field in the conductor connecting C1 and C2. Thus, in the series combination, charges on the two plates FIGURE 2.27 Combination of n (Β±Q) are the same on each capacitor. The total capacitors in series. 71 Reprint 2025-26 Physics potential drop V across the combination is the sum of the potential drops V1 and V2 across C1 and C2, respectively. Q Q + (2.55) V = V1 + V2 = C1 C 2 V 1 1 i.e., = + , (2.56) Q C1 C 2 Now we can regard the combination as an effective capacitor with charge Q and potential difference V. The effective capacitance of the combination is Q C = (2.57) V We compare Eq. (2.57) with Eq. (2.56), and obtain 1 1 1 = + (2.58) C C1 C 2 The proof clearly goes through for any number of capacitors arranged in a similar way. Equation (2.55), for n capacitors arranged in series, generalises to Q Q Q V = V1 + V 2 + ... + V n = + + ... + (2.59) C1 C 2 C n Following the same steps as for the case of two capacitors, we get the general formula for effective capacitance of a series combination of n capacitors: 1 1 1 1 1 = + + + ... + (2.60) C C1 C 2 C 3 C n 2.14.2 Capacitors in parallel Figure 2.28 (a) shows two capacitors arranged in parallel. In this case, the same potential difference is applied across both the capacitors. But the plate charges (Β±Q1) on capacitor 1 and the plate charges (Β±Q2) on the capacitor 2 are not necessarily the same: Q1 = C1V, Q2 = C2V (2.61) The equivalent capacitor is one with charge Q = Q1 + Q2 (2.62) and potential difference V. Q = CV = C1V + C2V (2.63) The effective capacitance C is, from Eq. (2.63), C = C1 + C2 (2.64) The general formula for effective capacitance C for parallel combination of n capacitors [Fig. 2.28 (b)] follows similarly, Q = Q1 + Q2 + ... + Qn (2.65) FIGURE 2.28 Parallel combination of i.e., CV = C1V + C2V + ... CnV(2.66) (a) two capacitors, (b) n capacitors. which gives C = C1 + C2 + ... Cn (2.67) 72 Reprint 2025-26 Electrostatic Potential and Capacitance Example 2.9 A network of four 10 mF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) FIGURE 2.29 Solution (a) In the given network, C1, C2 and C3 are connected in series. The effective capacitance CΒ’ of these three capacitors is given by 1 1 1 1 = + + C β² C1 C 2 C 3 For C1 = C2 = C3 = 10 mF, CΒ’ = (10/3) mF. The network has CΒ’ and C4 connected in parallel. Thus, the equivalent capacitance C of the network is  10 ο£Ά C = CΒ’ + C4 = ο£ο£¬ 3 + 10ο£Έο£· mF =13.3mF (b) Clearly, from the figure, the charge on each of the capacitors, C1, C2 and C3 is the same, say Q. Let the charge on C4 be QΒ’. Now, since the potential difference across AB is Q/C1, across BC is Q/C2, across CD is Q/C3 , we have Q Q Q + + = 500 V . C1 C 2 C 3 Also, QΒ’/C4 = 500 V. This gives for the given value of the capacitances, 10 β3 Q = 500 V Γ Β΅ F = 1.7 Γ 10 C and EXAMPLE 3 Q β² = 500 V Γ 10 Β΅ F = 5.0 Γ 10 β3 C 2.9
2.8 β In A Parallel Plate Capacitor With Air Between The Plates, Each Plate
Physics Class 11 Β· Chapter 2
2.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 Γ 10β3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? 79 Reprint 2025-26 Physics 2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. 2.10 A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? 2.11 A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Reprint 2025-26 Chapter Three CURRENT ELECTRICITY 3.1 INTRODUCTION In Chapter 1, all charges whether free or bound, were considered to be at rest. Charges in motion constitute an electric current. Such currents occur naturally in many situations. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results. The flow of charges in lightning is not steady, but in our everyday life we see many devices where charges flow in a steady manner, like water flowing smoothly in a river. A torch and a cell-driven clock are examples of such devices. In the present chapter, we shall study some of the basic laws concerning steady electric currents. 3.2 ELECTRIC CURRENT Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval t, let q+ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area. Similarly, let qβ be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+β qβ. This is proportional to t for steady current Reprint 2025-26 Physics and the quotient q I = (3.1) t is defined to be the current across the area in the forward direction. (If it turn out to be a negative number, it implies a current in the backward direction.) Currents are not always steady and hence more generally, we define the current as follows. Let DQ be the net charge flowing across a cross- section of a conductor during the time interval Dt [i.e., between times t and (t + Dt)]. Then, the current at time t across the cross-section of the conductor is defined as the value of the ratio of DQ to Dt in the limit of Dt tending to zero, βQ lim (3.2) I (t ) β‘ t 0 ββ βt In SI units, the unit of current is ampere. An ampere is defined through magnetic effects of currents that we will study in the following chapter. An ampere is typically the order of magnitude of currents in domestic appliances. An average lightning carries currents of the order of tens of thousands of amperes and at the other extreme, currents in our nerves are in microamperes. 3.3 ELECTRIC CURRENTS IN CONDUCTORS An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current. In nature, free charged particles do exist like in upper strata of atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules, a gram of water, for example, contains approximately 1022 molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials, the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied. In other materials, notably metals, some of the electrons are practically free to move within the bulk material. These materials, generally called conductors, develop electric currents in them when an electric field is applied. If we consider solid conductors, then of course the atoms are tightly bound to each other so that the current is carried by the negatively charged electrons. There are, however, other types of conductors like electrolytic solutions where positive and negative charges both can move. In our discussions, we will focus only on solid conductors so that the current is carried by the negatively charged electrons in the background of fixed positive ions. Consider first the case when no electric field is present. The electrons will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random. At a given time, there is no preferential 82 direction for the velocities of the electrons. Thus on the average, the Reprint 2025-26 Current Electricity number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. So, there will be no net electric current. Let us now see what happens to such a piece of conductor if an electric field is applied. To focus our thoughts, imagine the conductor in the shape of a cylinder of radius R (Fig. 3.1). Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and βQ put at the ends of a dielectric of the same radius and put of a metallic cylinder. The electrons will drift positive charge +Q distributed over one disc because of the electric field created to and similarly βQ at the other disc. We attach neutralise the charges. The current thus the two discs on the two flat surfaces of the will stop after a while unless the charges +Q cylinder. An electric field will be created and and βQ are continuously replenished. is directed from the positive towards the negative charge. The electrons will be accelerated due to this field towards +Q. They will thus move to neutralise the charges. The electrons, as long as they are moving, will constitute an electric current. Hence in the situation considered, there will be a current for a very short while and no current thereafter. We can also imagine a mechanism where the ends of the cylinder are supplied with fresh charges to make up for any charges neutralised by electrons moving inside the conductor. In that case, there will be a steady electric field in the body of the conductor. This will result in a continuous current rather than a current for a short period of time. Mechanisms, which maintain a steady electric field are cells or batteries that we shall study later in this chapter. In the next sections, we shall study the steady current that results from a steady electric field in conductors.
π Question Details
- Chapter
- Electrostatics
- Topic
- Parallel Plate Capacitor
- Year
- 2022
- Shift
- 29 Jul Shift 2
- Q Number
- Q11
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 2: Electrostatic Potential and Capacitance
More from this Chapter
Q43.A charged particle q is placed at the centre O of cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O. Then the electric flux through ABCD is (1) q/4ΟΟ΅0L (2) zero (3) q/2ΟΟ΅0L (4) q/3ΟΟ΅0L
Q44.If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is (1) Q/2 (2) βQ/2 (3) Q/4 (4) βQ/4
Q45.If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to (1) CV (2) 12 nCV2 (3) CV2 (4) 2n1 CV 2
Q46.Capacitance (in F ) of a spherical conductor with radius 1 m is (1) 1.1 Γ 10β10 (2) 10β6 (3) 9 Γ 10β9 (4) 10β3