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The charge q wouldn’t feel any electric force as medium of separation being metal because relative permitivity of metal is infinity so the force between Q and q is 0
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ASSUME THAT THE ENTIRE CHARGE OF THE TWO CONES IS CONCENTRATED AT THEIR CENTRE OF CHARGE(SAME AS CENTRE OF MASS) i.e DISTANCE BETWEEN THE TWO CHARGES IS 2x3H/4=3H/2 AND TOTAL CHARGE (MAGNITUDALLY) BE...
Dear student, The equipotential surface would be given by: For this surface to be equipotential, the potential is a constant, hence, So the correct option is option (c0. Regards Sumit
using A.M-G.M inequality: for maximum force between two charges q(Q-q) should be maximum accordingly q=Q-q ie 2q=Q or q/Q=1/2. WITH REGARDS SIDDHARTH GUPTA
for a simple pendulum, T=2#(l/g)^1/2 force experienced by charge=qE accl=F/m=qE/m , time period=2pi(mL/qE)^1/2
The dipole is pair of two equal and opposite charges so field due to it represents the vector sum of filed due to bothe the charges, but from one charge it is a single field.
To keep the first pith ball stationary, the net force acting on it should cancel aout to be zero. The force of gravity acting downwards= Mg=8*10/1000=0.08 N So, this should be the force acting on it...
The electric field is related to the potential as V= Ed. so, when the distance is doubled, So the electric field gets halved. The capacitance is given by So, when the distance is doubled So the...
The voltage across any two points is given by the work done in moving a charge particle from point A to B and dividing the work by the charge itself, i.e. the work done in moving a unit charge from...
2*10 11 electrons
Let r be the radius of small drops and big drop R. kq/r = 1 ….......(1) 1000* 4/3pi r^3 = 4/3piR^3 ….......(2) potential of big drop = kQ/R Substitute R in terms of r using eq 2 & find the ans
volume element, dv = 4*pi*r 2 *dr integration between 0 to R , ρ(r)dv =Q Integration [4pi*r 2 *dr A(R-r)]= Q Integration [4pi*r 2 *dr A(R-r)]= Q/4pi AIntegration [r 2 R-r 3 )dr] = A =...
Potential = kq1/r1+kq2/r2+kq3/r3 the net some give the total electrostatic energy. sher mohammad facultya askiitians btech. iit delhi
it will be zero as ,potential inside sphere is constant. sher mohammad faculty askiitians b.tech, iit delhi
*force on charge q is = F1-F2, F1 is repulsive force and F2 is attractive force due to negative charge. distance is s=3r s=ut-1/2*a*t^2 s+.5at^2=ut or 0=u-at sher mohammad faculty askiitians b.tech,...
Force between them is = kq1q2/r^2 < qE E> F/q sher mohammad faculty askiitians b.tech, iit delhi
vertical foce = mg horizontal force = qe tan (angle)= mg/qe tan(30)=mg/qe q= mg/tan(30)e sher mohammad faculty askiitians b.tech, iit delhi
There is force columbic force of attraction between the charged plates.
E.da =qin/e E.2pi*r*dl=λdL E=λ/2pir sher mohammad faculty askiitians b.tech, iit delhi