Chapter_04 - Solution.pdf - Chapter 4 Capacitors 1 If Q is...

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Chapter 4 Capacitors Your Turn 1 . If Q is doubled, the potential ( V ) will also double. C will not charge. It depends only on geometry of the conductor. 2. Q = CV = (2µF)(1000 V) = 2000 µC = 2mC Remember that farad X volt is coulomb. 3 . Solution:Potential of the two conductors is: V 1 = 1 1 20 2 Q C = = 10 V V 2 = 2 2 40 4 Q C = = 10 V Since they have equal potentials, no charge will flow between them when they are connected. No heat is dissipated. 4. If capacitance of B is C , then capacitance of A will be 2 C . After connection, the final common potential of the two spheres will be V = 1 2 20 20 2 3 Q Q C C C C C + µ + µ = + = 0 Hence, final charge on both the conductors will be zero. 5 . Final potential of the two spheres will be: V = 1 2 20 0 20 2 3 3 Q Q C V C C C C + µ + = = µ + q A = (2 C ) V = 40 3 C µ 6. E = V d = 3 50 1 10 × = 50,000 V/m. 7 . The electric field between the plates becomes 1 5 times. Therefore, the potential difference also becomes 1 5 times. 8 . C = 0 A d And 2C = ( ) 0 A x d x K + 2 0 A d = ( ) 0 3 A x d x + 2 3 x + 2( d – x ) = d d = 4 3 x x d = 3 4 9 . Refer to Example 6 for the reason explained in Example 6; the capacitance of the spherical capacitor will be C = 0 4 K ab b a π = 9 3 0.2 0.1 9 10 [0.2 01] × × × = 6.7 × 10 –11 F. 10 . Electric field between the plates is independent of distance between them. E = 0 σ Therefore, E will not change when distance between the plates is change. Energy density ( u = 1 2 ε 0 E 2 ) will not change 11 . Half the work done by the cell is lost as heat and remaining half gets stored in the capacitor. 12 . W cell = 100μJ Q · V = 100Μj (5μC) · V = 100μJ V = 20 V 13 . Charge on the capacitor after it is connected to the first cell is Q 1 = CV Work done by the first cell is W 1 = Q 1 V = CV 2 Charge on the capacitor after it is connected to the second cell is Q 2 = C (2 V ) = 2 CV Charge supplied by the second cell is = Q 2 Q 1 = CV Work done by the second cell is W 2 = ( CV )(2V) = 2 CV 2 2 1 2 2 1 2 2 W CV W CV = = Chapter_04 - Solution.indd 55 16-05-2019 19:30:22
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S.56 Electrostatics and Current Electricity 14 . Initial capacitance is C = 10µF Final capacitance is 2C = 20µF Initial charge is Q 1 = CV (10µF)(20V)=200µC Final charge is Q 2 = 2 CV = 400 µC W cell = ∆ QV = ( Q 2 Q 1 ) V = (200µC)(20V) = 4000 µJ = 4mJ Change in energy stored in the capacitor is U = U 2 – U 1 = 1 2 (2C) V 2 CV 2 = 1 2 CV 2 = 1 2 × (10µF)×(20V) 2 = 000 µJ = 2 mJ W extAgt + W cell = ∆ U or, W extAgt = 2 mJ – 4 mJ = –2mJ Electric field between the plates E = V d When d is halved, E doubles. Energy density, u = 2 0 1 2 E If E doubles, u becomes 4 times. 15 . Equivalent capacitance is given by 0 1 1 1 1 1 2 3 C = + + C 0 = 6 11 F µ Charge on each capacitor is same as charge supplied by the cell. Q = C 0 V = 6 11 F µ (66 V) = 36 µC Potential difference across the capacitors is V 1 = 1 36 1 Q C = = 36 V V 2 = 2 36 2 Q C = = 18 V V 3 = 3 36 2 Q C C F µ = µ = 12 V 16 . When two capacitors are connected in parallel, the equivalent capacitance is C 0 = C 1 + C 2 . When connected in series, the equivalent I less than both C 1 and C 2 . Therefore, parallel equivalent is 25µF and the series equivalent is 4µF (the smallest one). The two capacitances are 20µF and 5µF. C 1 + C 2 = 25µF and 1 2 1 2 20 5 4 20 5 C C F C C × = = µ + + 17 . In both the diagrams, the three capacitors are in parallel.
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