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Learn these 2 rules: Firstly For no translational motion, The force should act equal and oppsotely for no rotaion the net Torque should be zero on to the dipole.
Always use this method to solve the question similar to this: Initial charge across capacitor 100 pF, Q = CV => 100 x 10^-12 x 24 = > 24 x 10^10 C. Now when connected with other Capacitor 0f 20 pF...
r1 = 0.0529 x 1^2/1 f = k * e^2/r1 = > ke^2/0.0529 for 2 nd orbit, r2 = 0.0529 x 2^2/1) = > 0.0529 x 4 F =ke^2/(0.0529 x 4) Divide, F/f we get, => F/f = (ke^2/(0.0529 x 4)) *(0.0529/ke^2)) = > ¼ f =...
Reposting the answer plz check => Initial charge across capacitor 100 pF, Q = CV => 100 x 10^-12 x 24 = > 24 x 10^10 C. Now when connected with other Capacitor 0f 20 pF then assume charges Q and Q-x...
Initial charge across capacitor 100 pF, Q = CV => 100 x 10^-12 x 24 = > 24 x 10^10 C. Now when connected with other Capacitor 0f 20 pF then assume charges Q and Q-x in capicitors resecrvely. Thus,...
Force = K(Q-q)q/r^w Now Numerator = K(Q-q)q should be maximum Hence, q ahould be minium. Rherfore, dN/dq = k[Q-q-q] = 0 Q – 2q = 0 Q = 2q is tthe relation.
The special rubber tyre of air crafts are made slightly conducting so that electricity generated on account of friction b/w the tyres and the runway goes to earth.
Total flux = f1 + f2 = [k*2*pi*r*R/(R^2 + r^2)^(3/2)]*pi*r^2 + (kq/r^2)*pir^2 Wb/m^2 where f1 is total flusx due to Ring and f2 is the total flux due to f2.
ELECTRIC FIELD INTENSITY BETWEEN THE TWO PLATES = Q/2e0A = 100 Hence, Q/2Ae0 = 100 Q/Ae0 = 200 This would be the E in each plate.
Hello, Sorry i am not able to configure which question you are talking about? As your statement is incomplete and unsjustified. Plz repost the answer again.
E would be zero b/w both the chrges as both are psotive and would exhibits the repelling forces. Hence, let at x distance the Electric field would be due to both => E net = E1 + E2 = 0 20k/x^2 =...
Initially the bubble tends to remain in equilibrium with positive and negative charges respectively. Hence after adding some positive charge to the bubble the eqm. would be disturbed and hemce...
The answer would be: from work energy theorem, Electric Force x y = 0.5ky^2 Force = q*E Hence, y = 2qE/k meter. This would be the maximum xompression in spring. where E is the constant electric field.
from work energy theorem, Electric Force x y = 0.5ky^2 Force = q*E Hence, y = 2qE/k meter. This would be the maximum xompression in spring.
VERY EASY QUESTION. +q and -q makes a dipole. SEE ch electrostatics class12. suppose test charge on p. THEN due to +q it was repelled and due to -q it is attracted. then simple laws of vectors will...
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