Title 1.ELECTRIC CHARGES AND FIELDS - Explanations.pdf ✅

1 (b) When a dielectric K is introduced in a parallel plate condenser its capacity becomes K times. Hence C ′ = 5C0. Energy stored W0 = q 2 2C0 ∴ W′ = q 2 2C′ = q 2 2 × 5C0 ⇒ W′ = W0 5 2 (d) As the electron is moving in a straight line and starts from rest. Thus, from s = ut + 1 2 at 2 We get h = (0)t + 1 2 aet 2 t = √ 2h ae = √ 2hme eE (ae = eE me ) = √ 2 × 1.5 × 10−2 × 9.1 × 10−31 1.6 × 10−19 × 2 × 104 = 2.9 × 10−9 s 3 (c) E1 = ηq 4πε0a2 , E2 = ηq 4πε0a2 . Therefore E = E⃗ 1 + E⃗ 2 = √E1 2 + E2 2 + 2E1E2 cos 60° = √3ηq 4πε0a 2 Since η −1 < √3, 1 < √3η, √3η > 1 ⇒ √3ηq 4πε0a 2 > q 4πε0a 2 ⇒ E3 > E0 (E0 = q 4πε0a 2 ) 4 (c) When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to Q = CV, the charge will increase 5 (d) C = ε0A d ...(i) C ′ = ε0KA 2d ...(ii) From equation (i) and (ii) C ′ C = K 2 ⇒ 2 = K 2 ⇒ K = 4 6 (c) Top of the stratosphere has an electric field E is nearly equal to 100V/m 7 (d) Equivalent capacitance = 2×3 2+3 = 6 5 μF Total charge by Q = CV = 6 5 × 1000 = 1200μC Potential (V) across 2μF is V = Q C = 1200 2 = 600volt ∴ Potential on internal plates = 1000 − 600 = 400 V 8 (c) Initially charge on the capacitor Q = 10 × 12 = 120μC Finally charge on the capacitor Q ′ = (5 × 10) × 12 = 600μC So charge supplied by the battery later = Q ′ − Q = 480μC 9 (c) By Gauss’s law ε0φ = q φ = q ε0 = 1.8 × 10−6 8.85 × 10−12 = 2.0 × 105N − m2C −1 10 (c) Finitial = 1 4πε0 × 12×(−8) r 2 Finitial = 1 4πε0 × 96 r 2 Where ris the distance between them. When the charges are brought in contact, then q1 = q2 = 12−8 2 = 4 2 = 2μF ∴ Ffinal = 1 4πε0 × 2×2 r 2 = 4 r 2 × 1 4πε0 ⟹ |F| final = 4 r 2 × 1 4πε0
∴ |F| initial |F| final = 96 4 = 24 11 (d) Potential (V) = 3x 2 + 5 Intensity of the electric field= dV dx = 6x ∴ E at x = −2 = 6(−2) = −12Vm−1 12 (b) The situation is as shown in the figure Here, EA = σ 2ε0 and EB = σ 2ε0 Hence, the electric field between the plates E = EA + EB = σ 2ε0 + σ 2ε0 = σ ε0 and electric field outside the plates will be zero. 13 (c) During the discharge of a capacitor through a resistance charge at any instant Q = Q0e −t/CR ⇒ Q0 Q = e t/CR ⇒ t = CR loge Q0 Q If Q → constant, then t ∝ R Now, draw a line parallel to the time axis as shown. Suppose this line cut the graphs at points 1, 2 and 3. Corresponding times are t1,t2 and t3 respectively. Hence from graph t1 < t2 < t3 ⇒ R1 < R2 < R3 14 (d) Electric charge is quantised. It is an integral multiple of e = 1.60 × 10−19C 15 (b) F12 = 1 4πε0 . q 2 a2 and F13 = 1 4πε0 . q 2 (a√2) 2 ⇒ F12 F13 = 2 16 (a) C = 4πε0K [ ab b − a ] = 1 9 × 109 . 6 [ 12 × 9 × 10−4 3 × 10−2 ] = 24 × 10−11 = 240 pF 18 (c) FA = FB; because an uniform electric field produced between the plates 19 (b) New force F = 1 4πε0 (6−2)(2−2) r 2 × 10−12 = 0 20 (d) Minimum when connected in series and maximum when connected in parallel 21 (c) Electric field between sheets E = 1 2ε0 (σ − σ) = 0 23 (c) Electric field inside a conductor is always zero 24 (b) Ee = 1 2 Ea ∴ Ea = 2Ee 25 (b) If all charges are in equilibrium, system is also in equilibrium. Charge at centre : charge q is in equilibrium because no net force acting on it corner charge : If we consider the charge at corner B. This charge will experience following forces t Q R3 R2 R1 t1 t2 t3
FA = k Q 2 a2 , FC = kQ 2 a2 , FD = kQ 2 (a√2) 2 and FO = KQq (a√2) 2 Force at B away from the centre = FAC + FD = √FA 2 + FB 2 + FD = √2 kQ 2 a 2 + kQ 2 2a 2 = kQ 2 a 2 (√2 + 1 2 ) Force at B towards the centre = FO = 2kQq a2 For equilibrium of charge at B, FAC + FD = FO ⇒ kQ 2 a 2 (√2 + 1 2 ) = 2KQq a 2 ⇒ q = Q 4 (1 + 2√2) 26 (b) Potential at centre due to all charges are = 1 4πε0 [ q d + q d + q d + q d ] = 1 4πε0 4q d in S.I. unit = 4q d in C.G.S. unit 27 (a) The electric intensity outside a charged sphere. E = σR 2 ε0r 2 28 (a) Gravitational force FG = Gmemp r 2 FG = 6.7 × 10−11 × 9.1 × 10−31 × 1.6 × 10−27 (5 × 10−11) 2 = 3.9 × 10−47N Electrostatic force Fe = 1 4πε0 e 2 r 2 Fe = 9 × 109 × 1.6 × 10−19 × 1.6 × 10−19 (5 × 10−11) 2 = 9.22 × 10−8N So, Fe FG = 9.22×10−8 3.9×10−47 = 2.36 × 1039 30 (a) As there is no charge residing inside the cube, hence net flux is zero 31 (a) By using W = Q(E⃗ . ∆r ) ⇒ W = Q[(e1î+ e2ĵ+ e3k̂). (aî+ bĵ)] = Q(e1a + e2b) 32 (d) Electric field is directed right ward (higher potential of −200 V to lower potential of −400 V). When electron left free in an electric it accelerates opposite to the electric field. Hence in the given case electron accelerates left ward 34 (b) In balance condition ⇒ QE = mg ⇒ Q V d = ( 4 3 πr 3ρ)g ⇒ Q ∝ r 3 V ⇒ Q1 Q2 = ( r1 r2 ) 3 × V2 V1 ⇒ Q Q2 = ( r r/2 ) 3 × 600 2400 = 2 ⇒ Q2 = Q/2 36 (d) The two capacitors formed by the slabs may assumed to be in series combination 37 (a) By using E = 9 × 109 . 2pr (r 2−l 2)2 ; where p = (500 × 10−6) × (10 × 10−2) = 5 × 10−5 c × m, r = 25 cm = 0.25 m, l = 5 cm = 0.05 m O C B A D +q FO FC FD FAC FA a B– q +q 10 cm 20 cm 25 cm
E = 9 × 109 × 2 × 5 × 10−5 × 0.25 {(0.25) 2 − (0.05) 2} 2 = 6.25 × 107N/C 38 (a) Potential at mid point k(−q) d O, V = = 0 kq d + 39 (d) Charges on capacitors are Q1 = 30 × 2 = 60pC and Q2 = 20 × 3 = 60pC or Q1 = Q2 = Q (say) The situation is similar as the two capacitors in series are first charged with a battery of emf 50 V and then disconnected ∴ when S3 is closed V1 = 30V and V2 = 20V 40 (d) Potential difference across both the lines is same i.e. 2V. Hence charge flowing in line 2 Q = ( 2 × 2 2 + 2 ) = 2μC So charge on each capacitor in line (2) is 2 μC Charge time 1Q = 2 × 1 = 2μC 41 (a) Resolve E along CO and BO into two perpendicular components The sine components cancel each other The cosine components add up along OA to give 2E cos 60o ∴ Resultant field along AO = 2E − 2E cos 60° = 2E − E = E ∴ Resultant field is E along AO 42 (b) According to energy conservation, energy remains the same ⇒ Uparallel = Useries ⇒ 1 2 (nC)V 2 = 1 2 ( C n ) V ′2 ⇒ V ′ = nV [V ′ = potential difference across series combination] 43 (a) In steady state potential difference a cross capacitor = 2V So charge on capacitor Q = 10 × 2 = 20 μC 44 (d) ∵ charge 8q is placed at one corner of the cube, it can be imagined to be placed at the centre of a large cube which can be formed using 8 similar cubes and arranging them Now 8q is at centre of the 8 cubes arranged to form a closed box Total flux through the bigger cube = 8q ε0 [Gauss’s law] ⇒ Flux through one small cube = 1 8 × 8q ε0 = q ε0 45 (d) The electric field between the spheres of a charged capacitor is non-uniform and it decreases with distance from the Centre as E ∝ 1 r 2 46 (b) 1 C = 1 3 + 1 6 ⇒ C = 2 pF Total charge = 2 × 10−12 × 5000 = 10−8C The new potential when the capacitors are connected in parallel is 2pF Q Q 2pF 50 V + – + – + – + – + Q = 60 pC Q = 60 pC V1 = 30 V V2 = 20 V  2F 2F 1F Line (2) Line (1) 2V +q O – q d d 2d