12 Electrostatics. Target Publications Pvt. Ltd (1) Chapter 12: Electrostatics. q E = Hints to Problems for Practice

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1 hapter : Electrostatics Electrostatics Hints to Problems for Practice q. N or q N ε ε q µ q q. N or k εk NK k q 8.85 µ 8.85, l cm m, ε 8.85 /Nm q φ ε Nm /. q µ, r cm m q φ ε Nm / 5. φ E ds cos θ φ 8. cos θ.8.5 θ cos (.88). ds cm m, E N/ φ E ds cos θ φ cos 9 ( cos 9 ) φ cos 9 Wb ( cos ) φ cos (9 ).5 Wb 7. A m, q.5 µ.5, k (for air), ε 8.85 /Nm q E kaε.5 E N/. 8. µ/m /m, k E kε N/ 9. R. m, r m, λ.77 µ/m.77 /m, k Using, E kaε λ r 9.77 ().59 V/m. 9. q 9, l cm m, r cm m. k (for air) λ E πkεr q πε l kr V/m.. R mm m,.5 µ/m.5 /m, k., r m, R πr E kεr πεkr πr πε kr V/m 9

2 . q µ, r m q N. 7 ε 8.85 q E. πε r 9 9 () 9 5 V/m..885 µ/m.885 /m, k (for air), ε 8.85 /Nm f ε (.885 ) f N/m. ds 5m, q 8.85 µ 8.85, k q 8.85 ds /m F ds ε k ( ) N. 5. ds cm m, q. µ. Using, E ε q ds ε V/m Now using, f ε ( q/ds ) ε Std. XII Sci.: Physics Numericals (. / ) 8.85 ( 5 ) N/m. ds.5 m, q µ q ds.5 5 /m ds ( 5 F ).5 ε N ds F. ε k 5.5 N 9 q 9 q 7. V πεk R 5 R V R 5 5 We get, q µ q Now using, πr. () /m 5 ( 8.88 ) Using, f εk N/m 8. u εe 8.85 ( ).5 J/m u 8 J/m, k (for air) Using, u ε E u or E ε V/m 8

3 . q µ, r cm m q E πε r E 9 9 ( ) 5 5 V/m Now using, u ε E 8.85 (5 5 ) 89. J/m. d mm m, V V E V d 5 V/m u ε E 8.85 ( 5 ). u ε KE or E.5 J/m.5 J/m u ε k V/m. A. m, V V, k d.5 cm.5 m 5 m E V d E V/m u ε ke u 8.85 (8 5 ) J/m. V V, A cm m, d. mm. m m, E V d E V/m u ε E u 8.85 ( ) 8.85 hapter : Electrostatics 7.7 J/m. 5. d mm m, µf F Aε d or A d ε m. d c 8 cm 8 m, r c m, d s cm m r s m s Aε πε r s d πr c π d rc d r ( ) d m mm 7. D cm m R m, d.5 cm.5 m Aε πr ε d d. ( ) F 5 pf

4 Now using, πε R 5 5 R R.5 m. 8. µf F, d d, A A Aε d A ε d A ε d A d A d µf. 9. µf F, V 5 V, work done energy stored U V (5) J. 8 µf 8 F, V V, U V U 8 () J. µf F, E J Q E Q E Q. µf F, Q m Q U ( ) U J Std. XII Sci.: Physics Numericals. µf F, V V, k E V E ( ) 5 J. Now, k F V 5 V 5 V 5 V 5 V. V... V.. µf. F, V. kv, k 7.5 U V U. (. )... J Now using, U kv V U k V 5. AB µf + µf µf B µf + µf µf + + AD AD µf

5 . µf, µf and 5 µf are in parallel, µf Now, µf µf + µf.(i) Also, µf µf + µf Now, and µf are in series + µf.(ii) Finally, and are in parallel, From equation (i) and (ii), AB + 5 µf 7. µf µf eq + µf, V V Now, µf and µf are in series. 5 + F. µf 5 The charge acquired by the entire combination V 5 µ As eq and µf are in series, they have the same charge 5 µ 5 P.D. across µf 8 V As µf and µf are parallel, 5 V V V. harge across µf µf µ and charge across µf µf µ 8. µf F, V 5 V, µf F, Eq. + Eq. ( + ) 5. Now, charge across each capacitor remains the same in series. hapter : Electrostatics Q Q Eq. V Q 8 V 9 V Q 8 V V 9. Let and be the individual capacitances. S + and P + +. () and + 8. () 8 or 7 7 Substituting in equation () we get, ( ) ( ) or 8 or 8 The capacitances are µf and µf.. µf, 5 µf, eq µf Q V Q V Q V 5 5 U V U V (). J 5

6 U V 5 () J. eq + µf + µf µf Using, Q V we get, Q V 5 Q V Net charge, Q Q + Q Now using, V Q eq we get, ommon potential, V V. 5 µf, µf, µf, V V eq eq µf.857 µf 7 Net charge across the combination µ 7 As, and are in series, they have the same charge µ P.D. across 5 µf µ 5µ F V 5 P.D. across µf V P.D. across µf V 5 Std. XII Sci.: Physics Numericals. µf, µf V V, V eq µf Total charge Q + Q V + ommon potential, V. V 5. i. + + s s 8 µf. µf 7 q s V.. ii. P µf q V µ q V µ q V 8 8 µ 5. Side of cube 5 cm.5 m, k 8, E V/m, k u k E u 8 u J J Total energy u volume E u (.5). (.5).88 J. J Since, u k, E E k or E k k 8 E E. E.77 J.77 J

7 . A m, V V, d. cm m, k 7, ε Ak Now using, U V π 8.85 () 7 R.97 R.8 m hapter : Electrostatics Hints to Multiple hoice Questions. n Q e. 9. U () U J.79 J 7. P µf µf Q. F πε qq r 9 9. E.5 q.5 N/ 5 N/. E q πε r 9 ( ) 9. N/. 9 N X Initial capacitance, s µf s. µf 5 Final capacitance, s µf Now, ( + x) is in series with.. + s ( + x) + + x (+ x) + x. + x. x.x..9x x.. µf.9 µf R 8. E εkr From formula, R Eεkr 5. Volume of drop V πr Mass of oil drop volume density m vρ For stationary drop, mg q E mg v g r g E ρ π ρ q q e 7 π ( ) V/m. V/m. F k k F F k 7. N 5 N q.5 ε V m 8. The cube has six surfaces and as the charge is at its centre, it will produce equal number of lines of forces through each surface. Q The charge Q will produce in all lines ε of force. 7

8 Q Each surface will allow ε ds mm m, θ, dφ Eds cos Nm /. T.N.E.I. q R R π. E kε r πεkr.5. 9 E V/m q. E πε r q 9 q 5 µ 9. A. m, Q µ 5 Q /m A. Q 8 µ 8, R m.5 m K.8 9 N/m Volume stress Bulk Modulus (K) Volume strain F/ds K volume strain k ε(volumestrain) Volume strain kε K Q Q. πr kε K πr Q 9 π R ε.8 Std. XII Sci.: Physics Numericals ( ) 8 (.) (.5) ds Q 5. F ε εds (7.7 ) N..5 µ/m.5 /m df ds kε df ds [ k for air] 9.7 N/m 9.7 N/m 7. l 5 cm 5 m, q, k q q F ( ds l + l ) εk εkds εkl () 8.85 (5 )..5 9 N 8. f ε ke f ().5.5 N/m q q 9. f εk εkds εk(πr ) q...( ds πr ) εkπ R (8 ) f ( ) N/m

9 . E V/m, k u ε ke 8.85 ( ) 7.7 J/m. u ε E ε V R 8.85 u ( ) ( ). J/m. E V/m Volume ml m, u E ε 8.85 () Energy stored in m of air J. u ε ke E E u.5 ε. 7. W ε ke dv 5 k 8.85 N/ W 8.85 ( ) ( ) J 5. 8 J.5 7 J 5. x 8 x x 5 Using, u E u ' or x x u x 5x 5 or u u 5.. Q m, µf F V Q/ V hapter : Electrostatics 7. V V, d. cm. m, Q µf F E V/d V/m F QE 7.7 N 8. Q 5 µ, Q 5 µ, k V V Q k Q k k k 5 µ 5 µ k or k 9. Total charge Q Q + Q V + V 5 + ( + ) Effective capacitance p + µf + µf µf ommon potential difference Q V d. µf, d Using, we get, d d d d d / µf. r cm m, d. cm. m m, k, t m ε A and A πr t d t+ k 9

10 8.85. ( ) F pf. Q, V + V, V V Q V ( ).5 F. R km m π ε R µf. t. cm, apacity of capacitor ε Q V AV. () d After inserting a slab, capacitance becomes and charge remains same. Q V By increasing a distance, we get same potential difference as in first case. Q V. () d εa kεa d.. + εa kεa From () and (), or d d.. + εa εa kεa d d. +.. k. k d m, V V u ε V d u J/m. J/m k Std. XII Sci.: Physics Numericals. µf F, Q Q W 8. J 7. E V, E E E V E V V V E V E E 8. E 8 J,. µf. F E V E V 8. 5 V 5 V 9. µf F, Q µ Q/V or V Q 5 Q Energy ( ) E 5 J 5 5 J. µf F, V V, 5µF 5 F, V 8 V U V U V + V [ ] [ + ] J.7 J

11 . 8 µf 8 F, V 5 V, V 5 V Using, U V we get, Increase in energy V V ( V V ) + s s µf Now, + s hapter : Electrostatics 8 (5 5) J.. µf F, (V) 5 V Heat produced Electrical energy stored U V. ( ) (5).75.7 J XY +. µf Join M and N together. Similarly join P and Q. Then the given circuit gets modified as shown in the above diagram. eq µf + µf + µf µf 5. X Y X Y PQ QR PS RS This is balanced Wheatstone s bridge circuit. Therefore, there is no charge on capacitor which is connected in the middle branch (PQ). The equivalent circuit becomes µf µf P A µf µf µf B µf R P S µf s + s µf + µf µf.. From the circuit (I), equivalent capacitance of parallel capacitors P + 8 µf µf A µf µf µf B II µf Equivalent capacitance of series connection of µf and P is given by, 8 s + 8 µf 8 µf Similarly, the equivalent capacitance for circuit (II) is, 8 s µf Now s, s and are in parallel. From given condition, µf µf 7. On connecting X to P, µf capacitor is charged to a constant potential (E). As connection of X is switched over to Q, the total charge on µf capacitor is shared between µf and µf capacitors, which is of original charge. + 9 I

12 8. apacitance of first capacitor ( ) 5 µf 5 F and its voltage (V ) V; apacitance of the second capacitor ( ) 5 µf 5 F and its voltage (V ) V. We know that for parallel combination, the common potential (V), V + V + (5 ) + (5 ) (5 ) + (5 ) 5 V. 9. harge on capacitor Q V 5 5 µ Total capacity + Q 5µ Final P.D., V + 5µ + or µf Std. XII Sci.: Physics Numericals 5. Q V µ s µf 8 Q Now, V V V s Q V

Chapter 24: Capacitance and Dielectrics. Capacitor: two conductors (separated by an insulator) usually oppositely charged. (defines capacitance)

Chapter 24: Capacitance and Dielectrics. Capacitor: two conductors (separated by an insulator) usually oppositely charged. (defines capacitance) hapter 4: apacitance and Dielectrics apacitor: two conductors (separated by an insulator) usually oppositely charged a b - ab proportional to charge / ab (defines capacitance) units: F / pc4: The parallel