Tiêu đề 1C.ELECTROSTATICS(64-88).pdf ✅

Jr Chemistry E/M ELECTROSTATICS 64 NISHITH Multimedia India (Pvt.) Ltd., NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED VOL - VI LEVEL-VI HINTS 1. r dx  r dA r dr  2   dQ rdr   2 2 0  2 r dr 2 0 0 2 R Q r dr    3 0 2 3   R 2. r r r dr  dA r dr  2   3 dQ r dr  2 0 2 3 0 2 R R Q r dr      2 0 4 4 16 4 R R    4 0 15 2  R  3. dA r dr  2   dQ rdr   2 0 2 2 rdr r    0 2 dr r   2 0 2 R R dr Q r      2 0 2 ln R R   r 0  2 ln 2  4. dr r dq dv     2 2 0    r r dr 4 4 0  4 r dr 5 0 4 5 Q dq Q R      5. q q q q q Q F1 F2 F3 F2 300 F1 0 30 0 30 y x q q q q q Q F1 F2 F3 F2 300 F1 0 30 0 30 y x 1 2 3 4 2 sin 30 2 cos30 0 F F F F     2 2 2 2 2 2 2 3 2. 0 3 4 2 kq kq kq kQq a a a a     3 1 1 3 4 Q q           15 4 3 12 Q q         
NISHITH Multimedia India (Pvt.) Ltd., 65 ELECTROSTATICS NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED VOL - VI 6. 0 37 d Fcos37 mgsin37 0 37 N 2 KQq d F  mgcos37 Fsin37 mg 0 37 0 37 d Fcos37 mgsin37 0 37 N 2 KQq d F  mgcos37 Fsin37 mg 0 37 Friction can be either up or down the inclined plane If friction is up the plane 2 2 3 4 4 3 . 0.1 5 5 5 5 kQq KQq mg mg d d          2 4.3 0.4 3 KqQ mg mg d   2 4.3 2.6 KQq mg d  2 26 43 mgd q KQ  If the friction of down the plane 2 2 4 3 4 3 0.1 5 5 5 5 KQq mg mg KQq d d          2 2 4 0.3 3.4 KQq KQq mg d d   2 3.7 3.4 KQq mg d  2 34 37 mgd q KQ  7. + + + + + + ++ - - - - - - - - - - - dF    d d dF dF sin + + + + + + ++ - - - - - - - - - - - dF    d d dF dF sin F dF  2 sin    2 0 2 sin 4 R d q R        0 2 sin 4 q d R          /2 0 0 2 2 cos 4 Q q r R        2 2 0 Qq  R   8. Initially 1 2 2 0 1 4 q q F  r   If a dielectric slab of thickness ‘t’ is introduced the equivalent separatin between charges becomes kt r t       1 1 2 2 0 1 4 q q F kt r t           given 1 4 9 F F    2 2 1 4 1 9 r k t r t     2 2 1 4 9 . 2 2 r r r k         1 3 2 2 2 k    1, 2 2 4 k k k    9. r1  r3  r2  1 q 3 q 2 q A C B Y X O r1  r3  r2  1 q 3 q 2 q A C B Y X O The three charges must be on the same line 3 1 2 3 2 1 AC r r CB r r AB r r       , ,          Let 3 1 2 3 2 1 r r r r r r r AC CB AB             
Jr Chemistry E/M ELECTROSTATICS 66 NISHITH Multimedia India (Pvt.) Ltd., NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED VOL - VI   1 3 1 3 2 2 0 0 A q q q q F K r K r AB AC         2 3 q q AB AC   2 3 2 1 3 1 q q r r r r         similarly 0, FB   1 3 2 1 2 3 q q r r r r         0 FC   1 2 2 1 2 3 q q r r r r         using third equation q r r q r r 1 2 3 2 3 1            1 2 2 1 3 1 2 q r q r r q q       10. It is based on static equillibrium E D A C B F FR F 0 30 0 30 3cm 600 3cm  T T cos T sin mg 3F B E D A C B F FR F 0 30 0 30 3cm 600 E 3cm D A C B F FR F 0 30 0 30 3cm 600 3cm  T T cos T sin mg 3F B  T T cos T sin mg 3F B BE Tan AE   T F sin 3   T mg cos    3 1 F Tan mg    As per the given data ' '  is small sin BE Tan AB        AE cm   3 / 3 3  from 1 2 9 2 4 3 3 10 3 9 10 1 9 10 10 10 q           2 17 q 10    17/2 q 10    11. It is based on friction and colomb’s law 0 30 1 q 2 1m q 0 30 1 q 2 1m q In the first case 1 2 2 0 1 sin 4 q q mg f r      In second case 1 1 2 2 0 1 sin 4 q q mg f r        1 1 1 2 2 0 1 2 sin 4 q q q mg r       9 16 9 10 15 10 2 6 1 2 66.5 10 10 1 2 q            9 q C 2 5 10     12. It is based on dynamics and Coulombs Law x   mg T
NISHITH Multimedia India (Pvt.) Ltd., 67 ELECTROSTATICS NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED VOL - VI T mg cos  , 2 2 0 sin 4 q T x     2 2 0 4 q Tan x mg     , tan sin 2 x x l l       2 2 0 2 4 x q l  x mg    2 3 0 2 q l x  mg   2 0 2 3 2 dx ql dq x dt mg dt    dx v dt x    2 0 2 3 2 ql dq x x mg dt     3/2 0 2 3 2 ql dq x mg dt     0 0 2 3 2 2 l ql dq q mg mg dt       0 3 2 2 dq mg dt l      13. It is based on dynamics and Coulombs Law T A B q  l mg T l    2 0 1 4 Qq F  x   x T A B q  l mg T l    2 0 1 4 Qq F  x   x Applying Lamis theorem sin sin     F mg        But 2     2      2 2         sin cos 2 F mg     .2sin cos 2 2 cos 2 mg F      2 sin 2 F mg   2 0 1 2 sin 4 2 Qq mg x     substituting 2 sin 2 x l   2 2 0 1 2 sin 4 2 4 sin 2 Qq mg l      3 2 0 sin 2 32 Qq mgl     1 3 2 0 2sin 32 Qq mgl       14. It is based on fluids and coulombs law x  mg F 