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PhysicsMediumNumerical2023 · 11 Apr Shift 1

Q11.A parallel plate capacitor of capacitance 2 F is charged to a potential 𝑉. The energy stored in the capacitor is 𝐸1. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The 𝐸2 energy stored in the combination is 𝐸2. The ratio is 𝐸1 (1) 2: 1 (2) 2: 3 (3) 1: 2 (4) 1: 4

What This Question Tests

This question requires calculating the total energy contained in a given volume by first determining the average electric energy density of the electromagnetic wave.

Concepts Tested

Electric field amplitudeEnergy density of electromagnetic wavesEnergy in a volume

Formulas Used

u_E = (1/2)ε₀E²_rms

E_rms = E₀/√2

Energy = u_avg * V

📚 NCERT Sections This Tests

2.15Energy Stored In A Capacitor

Physics Class 11 · Chapter 2

82% match

2.15 ENERGY STORED IN A CAPACITOR A capacitor, as we have seen above, is a system of two conductors with charge Q and –Q. To determine the energy stored in this configuration, consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by 73 Reprint 2025-26 Physics bit, so that at the end, conductor 1 gets charge Q. By charge conservation, conductor 2 has charge –Q at the end (Fig 2.30 ). In transferring positive charge from conductor 2 to conductor 1, work will be done externally, since at any stage conductor 1 is at a higher potential than conductor 2. To calculate the total work done, we first calculate the work done in a small step involving transfer of an infinitesimal (i.e., vanishingly small) amount of charge. Consider the intermediate situation when the conductors 1 and 2 have charges Q¢ and FIGURE 2.30 (a) Work done in a small –Q¢ respectively. At this stage, the potential difference step of building charge on conductor 1 V¢ between conductors 1 to 2 is Q¢/C, where C is thefrom Q¢ to Q¢ + d Q¢. (b) Total work done in charging the capacitor may be capacitance of the system. Next imagine that a small viewed as stored in the energy of charge d Q¢ is transferred from conductor 2 to 1. Work electric field between the plates. done in this step (d W), resulting in charge Q¢ on conductor 1 increasing to Q¢+ d Q¢, is given by Q ′ δW = V ′δQ ′ = δQ ′ (2.68) C Integrating eq. (2.68) Q 2 Q 2 Q ′ 1 Q ′ Q δ Q ’ = = W = ∫ C 2 2C 0 C 0 We can write the final result, in different ways Q 2 1 2 1 W = = CV = QV (2.69) 2C 2 2 Since electrostatic force is conservative, this work is stored in the form of potential energy of the system. For the same reason, the final result for potential energy [Eq. (2.69)] is independent of the manner in which the charge configuration of the capacitor is built up. When the capacitor discharges, this stored-up energy is released. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates. To see this, consider for simplicity, a parallel plate capacitor [of area A (of each plate) and separation d between the plates]. Energy stored in the capacitor 1 Q 2 ( Aσ)2 d = = × (2.70) 2 C 2 ε0 A The surface charge density s is related to the electric field E between the plates, σ E = (2.71) ε0 From Eqs. (2.70) and (2.71) , we get Energy stored in the capacitor 2 × A d (2.72) U = (1/2 ) ε0 E 74 Reprint 2025-26 Electrostatic Potential and Capacitance Note that Ad is the volume of the region between the plates (where electric field alone exists). If we define energy density as energy stored per unit volume of space, Eq (2.72) shows that Energy density of electric field, u =(1/2)e0E 2 (2.73) Though we derived Eq. (2.73) for the case of a parallel plate capacitor, the result on energy density of an electric field is, in fact, very general and holds true for electric field due to any configuration of charges. Example 2.10 (a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system? FIGURE 2.31 Solution (a) The charge on the capacitor is Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C The energy stored by the capacitor is = (1/2) CV 2 = (1/2) QV = (1/2) × 9 × 10–8C × 100 V = 4.5 × 10–6 J (b) In the steady situation, the two capacitors have their positive EXAMPLE plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V¢. The 2.10 75 Reprint 2025-26 Physics charge on each capacitor is then Q¢ = CV¢. By charge conservation, Q¢ = Q/2. This implies V¢ = V/2. The total energy of the system is 1 1 − 6 = 2 × Q ' V ' = QV = 2.25 × 10 J 2 4 Thus in going from (a) to (b), though no charge is lost; the final energy is only half the initial energy. Where has the remaining energy gone? 2.10 There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in EXAMPLE the form of heat and electromagnetic radiation. SUMMARY 1. Electrostatic force is a conservative force. Work done by an external force (equal and opposite to the electrostatic force) in bringing a charge q from a point R to a point P is q(VP–VR), which is the difference in potential energy of charge q between the final and initial points. 2. Potential at a point is the work done per unit charge (by an external agency) in bringing a charge from infinity to that point. Potential at a point is arbitrary to within an additive constant, since it is the potential difference between two points which is physically significant. If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given is given by 1 Q V ( r ) = 4 πεo r 3. The electrostatic potential at a point with position vector r due to a point dipole of dipole moment p placed at the origin is 1 p.rˆ V ( r ) = 2 4 πεo r The result is true also for a dipole (with charges –q and q separated by 2a) for r >> a. 4. For a charge configuration q1, q2, ..., qn with position vectors r1, r2, ... rn, the potential at a point P is given by the superposition principle 1 q1 q 2 qn V = ( + + ... + ) 4 πε0 r1P r2P rnP where r1P is the distance between q1 and P, as and so on. 5. An equipotential surface is a surface over which potential has a constant value. For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces. The electric field E at a point is perpendicular to the equipotential surface through the point. E is in the direction of the steepest decrease of potential. Reprint 2025-26 Electrostatic Potential and Capacitance 6. Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. Potential energy of two charges q1, q2 at r1, r2 is given by 1 q1 q 2 U = 4 πε0 r12 where r12 is distance between q1 and q2. 7. The potential energy of a charge q in an external potential V(r) is qV(r). The potential energy of a dipole moment p in a uniform electric field E is –p.E. 8. Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by σ E = nˆ where ˆn is the unit vector along the outward normal to the ε0 surface and s is the surface charge density. Charges in a conductor can reside only at its surface. Potential is constant within and on the surface of a conductor. In a cavity within a conductor (with no charges), the electric field is zero. 9. A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. C is determined purely geometrically, by the shapes, sizes and relative positions of the two conductors. The unit of capacitance is farad:, 1 F = 1 C V –1. For a parallel plate capacitor (with vacuum between the plates), A C = ε0 d where A is the area of each plate and d the separation between them. 10. If the medium between the plates of a capacitor is filled with an insulating substance (dielectric), the electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarisation, gives rise to a field in the opposite direction. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. Consequently, the capacitance C increases from its value C0 when there is no medium (vacuum), C = KC0 where K is the dielectric constant of the insulating substance. 11. For capacitors in the series combination, the total capacitance C is given by 1 1 1 1 = + + + ... C C1 C 2 C 3 In the parallel combination, the total capacitance C is: C = C1 + C2 + C3 + ... where C1, C2, C3... are individual capacitances. 77 Reprint 2025-26 Physics 12. The energy U stored in a capacitor of capacitance C, with charge Q and voltage V is 1 1 2 1 Q 2 U = QV = CV = 2 2 2 C The electric energy density (energy per unit volume) in a region with electric field is (1/2)e0E2. Physical quantity Symbol Dimensions Unit Remark Potential or V [M1 L2 T–3 A–1] V Potential difference is physically significant Capacitance C [M–1 L–2 T–4 A2] F Polarisation P [L–2 AT] C m-2 Dipole moment per unit volume Dielectric constant K [Dimensionless] POINTS TO PONDER 1. Electrostatics deals with forces between charges at rest. But if there is a force on a charge, how can it be at rest? Thus, when we are talking of electrostatic force between charges, it should be understood that each charge is being kept at rest by some unspecified force that opposes the net Coulomb force on the charge. 2. A capacitor is so configured that it confines the electric field lines within a small region of space. Thus, even though field may have considerable strength, the potential difference between the two conductors of a capacitor is small. 3. Electric field is discontinuous across the surface of a spherical charged ˆn outside. Electric potential is, however shell. It is zero inside and σε0 continuous across the surface, equal to q/4pe0R at the surface. 4. The torque p × E on a dipole causes it to oscillate about E. Only if there is a dissipative mechanism, the oscillations are damped and the dipole eventually aligns with E. 5. Potential due to a charge q at its own location is not defined – it is infinite. 6. In the expression qV (r) for potential energy of a charge q, V (r) is the potential due to external charges and not the potential due to q. As seen in point 5, this expression will be ill-defined if V (r) includes potential 78 due to a charge q itself. Reprint 2025-26 Electrostatic Potential and Capacitance 7. A cavity inside a conductor is shielded from outside electrical influences. It is worth noting that electrostatic shielding does not work the other way round; that is, if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. EXERCISES

2.5A Parallel Plate Capacitor With Air Between The Plates Has A

Physics Class 11 · Chapter 2

82% match

2.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 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?

8.2A Parallel Plate Capacitor (Fig. 8.6) Made Of Circular Plates Each Of Radius

Physics Class 11 · Chapter 8

81% match

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26