Q61JEE Main 2023NATA parallel plate capacitor with air between the plate has a capacitance of 15pF15 {\text{pF}}15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes x/4pF\;{ {\text{x}}}/{4} {\text{pF}}x/4pF.The value of xxx is [24-Jan-2023 Shift 2]Show Answer🚀 Solve in Practice Mode📖 Explanation C0= ∈0A/d=15pF\; {\text{C}}_0=\;{\in _0 {\text{A}}}/{ {\text{d}}}=15 {\text{pF}}C0=∈0A/d=15pF C= K∈0A/2d= 3.52×15pF= 1054pF\; {\text{C}}=\;{ {\text{K}} \in _0 {\text{A}}}/{2 {\text{d}}}=\;\frac{3.5}{2} \times 15 {\text{pF}}=\;\frac{105}{4} {\text{pF}}C=K∈0A/2d=23.5×15pF=4105pF
Q62JEE Main 2023MCQThe distance between two plates of a capacitor is ddd and its capacitance is C1C_1C1, when air is the medium between the plates. If a metal sheet of thickness 2d/3\;{2 {\text{d}}}/{3}2d/3 and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes C2{\text{C}}_2C2. The ratio C2/C1\;{ {\text{C}}_2}/{ {\text{C}}_1}C2/C1 is [10-Apr-2023 shift 2]A. 4:14: 14:1B. 3:13: 13:1C. 2:12: 12:1D. 1:11: 11:1Show Answer🚀 Solve in Practice Mode📖 Explanationt= 2d/3{\text{t}}=\;{2 {\text{d}}}/{3}t=2d/3 K=∞{\text{K}}=\infty K=∞ for metalsC1= ϵ0A/d {\text{C}}_1=\;{\epsilon _0 {\text{A}}}/{ {\text{d}}}\;\;\;C1=ϵ0A/d C2= ϵ0A/d−t+ t/k= ϵ0A/d− 2d/3+0= 3ϵ0A/d\;\;\; {\text{C}}_2=\;{\epsilon _0 {\text{A}}}/{ {\text{d}}- {\text{t}}+\;{ {\text{t}}}/{ {\text{k}}}}=\;{\epsilon _0 {\text{A}}}/{ {\text{d}}-\;{2 {\text{d}}}/{3}+0}=\;{3 \epsilon _0 {\text{A}}}/{ {\text{d}}}C2=ϵ0A/d−t+t/k=ϵ0A/d−2d/3+0=3ϵ0A/d = C2/C1= 3ϵ0A/d/ϵ0A/d= 31=\;{ {\text{C}}_2}/{ {\text{C}}_1}=\;{\;{3 \epsilon _0 {\text{A}}}/{ {\text{d}}}}/{{\epsilon _0 {\text{A}}}/{ {\text{d}}}}=\;\frac{3}{1}=C2/C1=3ϵ0A/d/ϵ0A/d=13
Q63JEE Main 2022NATA parallel plate capacitor with width 4cm4 {\text{cm}}4cm, length 8cm8 {\text{cm}}8cm and separation between the plates of 4mm4 {\text{mm}}4mm is connected to a battery of 20V20 {\text{V}}20V. A dielectric slab of dielectric constant 5 having length 1cm1 {\text{cm}}1cm, width 4cm4 {\text{cm}}4cm and thickness 4mm4 {\text{mm}}4mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be _______ E0JE_0 {\text{J}}E0J. (Where E0E_0E0 is the permittivity of free space) [27-Jul-2022-Shift-2]Show Answer🚀 Solve in Practice Mode📖 ExplanationCeff=[ ϵ0(7×4)4/10+ 5ϵ0(1×4)4/10]×10−2{\text{C}}_{ {\text{eff}}}=[\;\frac{\epsilon _0(7 \times 4)}{4 / 10}+\;\frac{5 \epsilon _0(1 \times 4)}{4 / 10}] \times 10^{-2}Ceff=[4/10ϵ0(7×4)+4/105ϵ0(1×4)]×10−2 Ceff=1.2ϵ0{\text{C}}_{ {\text{eff}}}=1.2 \epsilon _0Ceff=1.2ϵ0Energy = 12CeffV2=\;\frac{1}{2} {\text{C}}_{ {\text{eff}}} {\text{V}}^2=21CeffV2 = 12(1.2)ϵ0(20)(20)=240ϵ0=\;\frac{1}{2}(1.2) \epsilon _0(20)(20)=240 \epsilon _0=21(1.2)ϵ0(20)(20)=240ϵ0
Q64JEE Main 2022NATThe equivalent capacitance between points AAA and BBB in below shown figure will be _______μF\mu {\text{F}}μF [25-Jun-2022-Shift-1]Show Answer🚀 Solve in Practice Mode📖 ExplanationCeq= (3×8)×8(3×8)+8C_{e q}=\;\frac{(3 \times 8) \times 8}{(3 \times 8)+8}Ceq=(3×8)+8(3×8)×8 = 24×832=\;\frac{24 \times 8}{32}=3224×8 =6μF=6 \mu {\text{F}}=6μF
Q65JEE Main 2022NATA capacitor C1{\text{C}}_1C1 of capacitance 5μF5 \mu {\text{F}}5μF is charged to a potential of 30V30 {\text{V}}30V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C2{\text{C}}_2C2 of capacitance 10μF10 \mu {\text{F}}10μF as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C2{\text{C}}_2C2 is ____μC\mu {\text{C}}μC [28-Jun-2022-Shift-2]Show Answer🚀 Solve in Practice Mode📖 ExplanationLet the charge qqq is flown in the circuit.So using Kirchhoff's law, q10= 150−q5\;\frac{q}{10}=\;\frac{150-q}{5}10q=5150−q q=100μCq=100 \mu Cq=100μC
Q66JEE Main 2022MCQThe charge on capacitor of capacitance 15μF15 \mu {\text{F}}15μF in the figure given below is : [26-Jun-2022-Shift-2]A. 60μc60 \mu {\text{c}}60μcB. 130μc130 \mu {\text{c}}130μcC. 260μc260 \mu {\text{c}}260μcD. 585μc585 \mu {\text{c}}585μcShow Answer🚀 Solve in Practice Mode📖 ExplanationCeq= 12026μFC_{e q}=\;\frac{120}{26} \mu FCeq=26120μF ⇒Q flown or Q= 13×12026μC=60μC\Rightarrow Q_{\;\text{flown }\;} \;\text{ or }\; Q=\;\frac{13 \times 120}{26} \mu C=60 \mu C⇒Qflown or Q=2613×120μC=60μC ⇒ Charge on 15μF capacitor =60μC\Rightarrow \;\text{ Charge on }\; 15 \mu {\text{F}} \;\text{ capacitor }\;=60 \mu {\text{C}}⇒ Charge on 15μF capacitor =60μCAs all the capacitors are in series.
Q67JEE Main 2022NATA parallel plate capacitor is made up of stair like structure with a plate area A{\text{A}}A of each stair and that is connected with a wire of length b{\text{b}}b, as shown in the figure. The capacitance of the arrangement is x15 E0Ab\;\frac{x}{15} \;\frac{E_0 A}{b}15xbE0A. The value of x{\text{x}}x is_____ [27-Jun-2022-Shift-2]Show Answer🚀 Solve in Practice Mode📖 ExplanationThe circuit is equivalent to 3 capacitors in parallel as shown Ceq= ϵ0Ab(1+ 13+ 15)= 2315 ϵ0AbC_{e q}=\;\frac{\epsilon _0 A}{b} (1+\;\frac{1}{3}+\;\frac{1}{5}) =\;\frac{23}{15} \;\frac{\epsilon _0 A}{b}Ceq=bϵ0A(1+31+51)=1523bϵ0A ⇒x=23\Rightarrow x=23⇒x=23
Q68JEE Main 2022MCQTwo capacitors having capacitance C1{\text{C}}_1C1 and C2{\text{C}}_2C2 respectively are connected as shown in figure. Initially, capacitor C1{\text{C}}_1C1 is charged to a potential difference V{\text{V}}V volt by a battery. The battery is then removed and the charged capacitor C1{\text{C}}_1C1 is now connected to uncharged capacitor C2{\text{C}}_2C2 by closing the switch S{\text{S}}S. The amount of charge on the capacitor C2{\text{C}}_2C2, after equilibrium, is : [26-Jun-2022-Shift-1]A. C1C2(C1+C2)V\;\frac{C_1 C_2}{(C_1+C_2)} V(C1+C2)C1C2VB. (C1+C2)C1C2V\;\frac{(C_1+C_2)}{C_1 C_2} VC1C2(C1+C2)VC. (C1+C2)V(C_1+C_2) V(C1+C2)VD. (C1−C2)V(C_1-C_2) V(C1−C2)VShow Answer🚀 Solve in Practice Mode📖 ExplanationV common = C1VC1+C2V_{\;\text{common }\;}=\;\frac{C_1 V}{C_1+C_2}Vcommon =C1+C2C1V ⇒\Rightarrow ⇒ Charge on capacitor C2C_2C2
Q69JEE Main 2022NATA composite parallel plate capacitor is made up of two different dielectric materials with different thickness (t1.(t_1.(t1. and .t2).t_2).t2) as shown in figure. The two different dielectric materials are separated by a conducting foil F{\text{F}}F. The voltage of the conducting foil is V. [26-Jul-2022-Shift-1]Show Answer🚀 Solve in Practice Mode📖 Explanation C1C2= 3×t2t1×4= 32\;\frac{C_1}{C_2}=\;\frac{3 \times t_2}{t_1 \times 4}=\;\frac{3}{2}C2C1=t1×43×t2=23 qC1=v1, qC2=v2\;\frac{q}{C_1}=v_1, \;\frac{q}{C_2}=v_2C1q=v1,C2q=v2 v1v2= C2C1= 23\;\frac{v_1}{v_2}=\;\frac{C_2}{C_1}=\;\frac{2}{3}v2v1=C1C2=32
Q70JEE Main 2022MCQA condenser of 2μF2 \mu {F}2μF capacitance is charged steadily from 0 to 5C5 {C}5C. Which of the following graph represents correctly the variation of potential difference (V)( {V})(V) across it's plates with respect to thecharge (Q)(Q)(Q) on the condenser? [25-Jul-2022-Shift-1]A.B.C.D.Show Answer🚀 Solve in Practice Mode📖 ExplanationQ=CVQ=C VQ=CVAs capacitance is constant Q∝V{Q} \propto {V}Q∝V and Vf= QfC= 52×10−6=2.5×106VV_{f}=\;\frac{Q_{f}}{C}=\;\frac{5}{2 \times 10^{-6}}=2.5 \times 10^{6} {V}Vf=CQf=2×10−65=2.5×106VSo correct graph will be A.
Q71JEE Main 2022MCQA source of potential difference VVV is connected to the combination of two identical capacitors as shown in the figure. When key ' KKK ' is closed, the total energy stored across the combination is E1E_1E1. Now key ' KKK ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E2E_2E2. The ratio E1/E2E_1 / E_2E1/E2 will be : [26-Jul-2022-Shift-2]A. 110\;\frac{1}{10}101B. 25\;\frac{2}{5}52C. 513\;\frac{5}{13}135D. 526\;\frac{5}{26}265Show Answer🚀 Solve in Practice Mode📖 Explanation(1) Switch is closedCeq−2C{\text{C}}_{ {\text{eq}}}-2 {\text{C}}Ceq−2CEnergy E1= 12CeqV2{\text{E}}_1=\;\frac{1}{2} {\text{C}}_{ {\text{eq}}} {\text{V}}^2E1=21CeqV2 = 122C×V2=\;\frac{1}{2} 2 {\text{C}} \times {\text{V}}^2=212C×V2 E1=CV2{\text{E}}_1= {\text{CV}}^2E1=CV2(ii) When switch is opened charge on right capacitor remain CV while potential on left capacitor remain sameDielectric K=5{\text{K}}=5K=5 C′=KC{\text{C}}^{'}= {\text{KC}}C′=KC C′=5CC^{'}=5 CC′=5C E2= 12(5C)V2+ (CV)2/2(5C){\text{E}}_2=\;\frac{1}{2}(5 {\text{C}}) {\text{V}}^{2+}\;{( {\text{CV}})^2}/{2(5 {\text{C}})}E2=21(5C)V2+(CV)2/2(5C) E2= 5CV2/2+ CV2/10{\text{E}}_2=\;{5 {\text{CV}}^2}/{2}+\;{ {\text{CV}}^2}/{10}E2=5CV2/2+CV2/10 E2= 13CV2/5{\text{E}}_2=\;{13 {\text{CV}}^2}/{5}E2=13CV2/5 E1E2= CV213CV25= 513\;\frac{E_1}{E_2}=\;\frac{C V^2}{\frac{13 C V^2}{5}}=\;\frac{5}{13}E2E1=513CV2CV2=135 E1/E2= 513\;{ {\text{E}}_1}/{ {\text{E}}_2}=\;\frac{5}{13}E1/E2=135
Q72JEE Main 2022MCQTwo identical thin metal plates has charge q1q_1q1 and q2q_2q2 respectively such that q1>q2q_1 > q_2q1>q2. The plates were brought close to each other to form a parallel plate capacitor of capacitance C{\text{C}}C. The potential difference between them is: [29-Jul-2022-Shift-2]A. (q1+q2)C\;\frac{(q_1+q_2)}{C}C(q1+q2)B. (q1−q2)C\;\frac{(q_1-q_2)}{C}C(q1−q2)C. (q1−q2)2C\;\frac{(q_1-q_2)}{2 C}2C(q1−q2)D. 2(q1−q2)C\;\frac{2(q_1-q_2)}{C}C2(q1−q2)Show Answer🚀 Solve in Practice Mode📖 ExplanationCharge on the left surface of plate A= Total charge 2{\text{A}}=\;\frac{\;\text{ Total charge }\;}{2}A=2 Total charge = q1+q22=\;\frac{q_1+q_2}{2}=2q1+q2Let right surface of plate A{\text{A}}A has charge =x=x=xAnd total charge on plate A=q1A=q_1A=q1 ∴q1=\therefore q_1=∴q1= Charge on left surface of plate A+{\text{A}}+A+ Charge on right surface of plate A{\text{A}}A = q1+q22+x=\;\frac{q_1+q_2}{2}+x=2q1+q2+x ⇒x=q1− q1+q22\Rightarrow x=q_1-\;\frac{q_1+q_2}{2}⇒x=q1−2q1+q2 = 2q1−q1−q22=\;\frac{2 q_1-q_1-q_2}{2}=22q1−q1−q2 = q1−q22=\;\frac{q_1-q_2}{2}=2q1−q2Let potential difference between two plates =v=v=v For capacitor we know,q=CVq=C Vq=CV ∴ q1−q22=CV\therefore \;\frac{q_1-q_2}{2}=C V∴2q1−q2=CV ⇒V= q1−q22C\Rightarrow V=\;\frac{q_1-q_2}{2 C}⇒V=2Cq1−q2
Q73JEE Main 2022NATA capacitor of capacitance 50pF50 {\text{pF}}50pF is charged by 100V100 {\text{V}}100V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is____ nJ. [27-Jun-2022-Shift-1]Show Answer🚀 Solve in Practice Mode📖 ExplanationElectrical energy lost = 12( 12CV2)=\;\frac{1}{2} (\;\frac{1}{2} C V^2)=21(21CV2) = 12× 12×50×10−12×(100)2=\;\frac{1}{2} \times \;\frac{1}{2} \times 50 \times 10^{-12} \times (100)^2=21×21×50×10−12×(100)2 = 5004nJ=\;\frac{500}{4} {\text{nJ}}=4500nJ =125nJ=125 {\text{nJ}}=125nJ
Q74JEE Main 2022MCQC0C_0C0 is the capacitance of a parallel plate capacitor with air as a medium between the plates (as shown in Fig. 1). If half space between the plates is filled with a dielectric of relative permittivity ϵr\epsilon _rϵr (as shown in Fig. 2), the new capacitance of the capacitor will be : [30-Jun-2022-Shift-1]A. Co2(1+ϵr)\;\frac{C_o}{2}(1+\epsilon _r)2Co(1+ϵr)B. Co+ϵrC_o+\epsilon _rCo+ϵrC. Coϵr2\;\frac{C_o \epsilon _r}{2}2CoϵrD. Co(1+ϵr)C_o(1+\epsilon _r)Co(1+ϵr)Show Answer🚀 Solve in Practice Mode📖 ExplanationCapacitance of Fig. 1, C0= ϵ0AdC_0=\;\frac{\epsilon _0 A}{d}C0=dϵ0AIn Fig. 2, C1{\text{C}}_1C1 and C2{\text{C}}_2C2 are parallel with each other,∴\therefore ∴ Equivalent capacitance,Ceq=C1+C2C_{e q}=C_1+C_2Ceq=C1+C2 = ϵ0A/2d+ ϵ0ϵrA/2d=\;\frac{\epsilon _0 A / 2}{d}+\;\frac{\epsilon _0 \epsilon _r A / 2}{d}=dϵ0A/2+dϵ0ϵrA/2 = ϵ0A2d+ ϵ0ϵrA2d=\;\frac{\epsilon _0 A}{2 d}+\;\frac{\epsilon _0 \epsilon _r A}{2 d}=2dϵ0A+2dϵ0ϵrA = Aϵ02d(1+ϵr)= C02(1+ϵr)=\;\frac{A \epsilon _0}{2 d} (1+\epsilon _r) =\;\frac{C_0}{2} (1+\epsilon _r)=2dAϵ0(1+ϵr)=2C0(1+ϵr)
Q75JEE Main 2022MCQA slab of dielectric constant K{\text{K}}K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 34d\;\frac{3}{4} {\text{d}}43d, where ddd is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :(Given C0={\text{C}}_0=C0= capacitance of capacitor with air as medium between plates.) [28-Jul-2022-Shift-2]A. 4KC03+K\;\frac{4 K C_0}{3+K}3+K4KC0B. 3KC03+K\;\frac{3 K C_0}{3+K}3+K3KC0C. 3+K4KC0\;\frac{3+K}{4 K C_0}4KC03+KD. K4+K\;\frac{K}{4+K}4+KKShow Answer🚀 Solve in Practice Mode📖 ExplanationC0= ϵ0AdC_0=\;\frac{\epsilon _0 A}{d}C0=dϵ0A C= ϵ0Ad− 3d4+ 3d4K= 4ϵ0AK3d+KdC=\;\frac{\epsilon _0 A}{d-\;\frac{3 d}{4}+\;\frac{3 d}{4 K}}=\;\frac{4 \epsilon _0 A K}{3 d+K d}C=d−43d+4K3dϵ0A=3d+Kd4ϵ0AK = 4KC03+K=\;\frac{4 K C_0}{3+K}=3+K4KC0
Q76JEE Main 2022MCQA parallel plate capacitor with plate area A{\text{A}}A and plate separation d=2m{\text{d}}=2 {\text{m}}d=2m has a capacitance of 4μF4 \mu {\text{F}}4μF. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K=3{\text{K}}=3K=3 (as shown in figure) will be [26-Jun-2022-Shift-2]A. 2μF2 \mu {\text{F}}2μFB. 32μF32 \mu {\text{F}}32μFC. 6μF6 \mu {\text{F}}6μFD. 8μF8 \mu {\text{F}}8μFShow Answer🚀 Solve in Practice Mode📖 ExplanationNow, ϵ0Ad=4μF\;\frac{\epsilon _0 A}{d}=4 \mu Fdϵ0A=4μF ⇒ 128 ϵ0Ad=6μF\Rightarrow \;\frac{12}{8} \;\frac{\epsilon _0 A}{d}=6 \mu F⇒812dϵ0A=6μF
Q77JEE Main 2022MCQIf the charge on a capacitor is increased by 2C2 {\text{C}}2C, the energy stored in it increases by 44%44 \%44%. The original charge on the capacitor is (in C) [24-Jun-2022-Shift-2]A. 10B. 20C. 30D. 40Show Answer🚀 Solve in Practice Mode📖 ExplanationLet initially the charge is q so 12 q2C=Ui\;\frac{1}{2} \;\frac{q^2}{C}=U_i21Cq2=UiAnd 12 (q+2)2C=Uf\;\frac{1}{2} \;\frac{(q+2)^2}{C}=U_f21C(q+2)2=UfGiven Uf−UiUi×100=44\;\frac{U_f-U_i}{U_i} \times 100=44UiUf−Ui×100=44 (q+2)2−q2q=.44\;\frac{(q+2)^{2-}q^2}{q}=.44q(q+2)2−q2=.44 ⇒q=10C\Rightarrow q=10 C⇒q=10C
Q78JEE Main 2022MCQA force of 10N10 {\text{N}}10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be. [27-Jun-2022-Shift-1]A. 5N5 {\text{N}}5NB. 10N10 {\text{N}}10NC. 20N20 {\text{N}}20ND. ZeroShow Answer🚀 Solve in Practice Mode📖 ExplanationE between two plates is σϵ0\;\frac{\sigma }{\epsilon _0}ϵ0σ and due to one plate is σ2ϵ0\;\frac{\sigma }{2 \epsilon _0}2ϵ0σ so the force will be halvedSo new force F=5N{\text{F}}=5 {\text{N}}F=5N
Q79JEE Main 2022MCQTwo capacitors, each having capacitance 40μF40 \mu {\text{F}}40μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K{\text{K}}K such that the equivalence capacitance of the system became 24μF24 \mu {\text{F}}24μF. The value of K{\text{K}}K will be : [28-Jul-2022-Shift-1]A. 1.51.51.5B. 2.52.52.5C. 1.21.21.2D. 3Show Answer🚀 Solve in Practice Mode📖 ExplanationC eq = C(KC)/C+KC= KC/K+1{\text{C}}_{\;\text{eq }\;}=\;{ {\text{C}}( {\text{KC}})}/{ {\text{C}}+ {\text{KC}}}=\;{ {\text{KC}}}/{ {\text{K}}+1}Ceq =C(KC)/C+KC=KC/K+1 24= K40/K+124=\;{ {\text{K}} 40}/{ {\text{K}}+1}24=K40/K+1 [K=1⋅5]{[ {\text{K}}=1 \cdot 5] }[K=1⋅5]
Q80JEE Main 2022MCQCapacitance of an isolated conducting sphere of radius R1{R}_{1}R1 becomes n{n}n times when it is enclosed by a concentric conducting sphere of radius R2R_{2}R2 connected to earth. The ratio of their radii ( R2R1)(\;\frac{R_{2}}{R_{1}})(R1R2) is : [25-Jul-2022-Shift-2]A. nn−1\;\frac{n}{n-1}n−1nB. 2n2n+1\;\frac{2 n}{2 n+1}2n+12nC. n+1n\;\frac{n+1}{n}nn+1D. 2n+1n\;\frac{2 n+1}{n}n2n+1Show Answer🚀 Solve in Practice Mode📖 ExplanationInitially =C0=4πϵ0R1=C_{0}=4 \pi \epsilon _{0} R_{1}=C0=4πϵ0R1Finally 4πϵ0R1R2R2−R1=nC0=4πϵ0nR1\;\frac{4 \pi \epsilon _{0} R_{1} R_{2}}{R_{2}-R_{1}}=n C_{0}=4 \pi \epsilon _{0} n R_{1}R2−R14πϵ0R1R2=nC0=4πϵ0nR1 R2R2−R1=n\;\frac{R_{2}}{R_{2}-R_{1}}=nR2−R1R2=n 1− R1R2= 1n1-\;\frac{R_{1}}{R_{2}}=\;\frac{1}{n}1−R2R1=n1 R1R2= n−1n\;\frac{R_{1}}{R_{2}}=\;\frac{n-1}{n}R2R1=nn−1 R2R1= nn−1\;\frac{R_{2}}{R_{1}}=\;\frac{n}{n-1}R1R2=n−1n