89 - Electricity Questions Answers
a fixed uniformly charged ring of radius 3m has a positive linear charge density 50/3µC/m. a point charge 5μC is moving towards the ring along its axis such that its kinetic energy 4m away is 5J. Its kinetic energy at the centre of the ring will be?
compare total energy at the two points and get answer
charge flowing thru a xsection of wire varies with time as q= 2te-kt , where k is const . the voltmeter reads zero at t=??
q= 2te-kt
on differentiating with respect to t, we get
dq/dt = 2[t(-k)e-kt + e-kt ]
so for V = 0, i will be 0
so t = 1/k
the effective power of 3 lamps A (100W , 200V) , B(60W,200V) ,C (40W,200V) connected in series is equal to 600/31W . how??
for series R = R1 + R2 + R3 + R4
and R = V2/P { This is the general case to find the resistence of the device)
now use this and get the result
a point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. due to induced charges on the sphere , find the elecric potential at point P on the surface of sphere in volt if kq/r =18V.
in the above ques, last time u found out potential at center but according to ques we hav eto find out on the surface and ans. is 3
if the answer is 3 V then 2r should be the distance of q from surface not from centre.
first confirm
In the ques previously submitted by someone which was as follow
the electric field vector is given by E= a(x)1/2 i^ . find the φ thru a cube bounded by surface x=l, x=2l, y=0, y=l , z=l, z=0.
Sir , my query is that in the solutn submitted the leaving flux at x=2l is given as( 2l)1/2 X l2 . but as the electric field is itself along x then why 2l will be used as along x axis it will be zero because cosθ used is perpendicular to surface with electric field axis thus cos90 =0??
θ should be measured between E and perpendicular drawn on the surface
4 particles each of mass m and charge q are held at the corners of a square of side a . they are released at t=0 and move under mutual repulsice force. find the speed of any particle when its distance from the center is double.
potential energy in initial condition = 4*k*q*q/a + 2*k*q*q/a√2
potential energy in final condition = 4*k*q*q/2a + 2*k*q*q/2a√2
and kinetic energy in the final condition = 4*mv2/2
now use energy conservation
a point charge q is placed at a distance 2r from the centre O of a conducting charged sphere of radius r. due to induced charges on the sphere , find the elecric potential at point P on the surface of sphere in volt if kq/r =18V.
potential at the centre due to q = kq/2r = 9 V
potential at centre due to surface charge = 0 V
total potential = 9 V
now at surface potential due to q = kq/r = 18 V
but it should be 9 V totally so due to surface induction, potential will be -9V
the electric field vector is given by E=a(x)½ iˆ.find the flux through a cube bounded by surfaces x=l ,x=2l, y=0 ,y=l, z=0, z=l
entering flux at x=l will be a(l)1/2 * l2
leaving flux at x=2l will be a(2l)1/2 * l2
so net flux will be the difference of these 2 flux
What would be the Potential energy of two points charges q &-q which r separated by a dist d ( IF POTENTIAL AT MID PT IS TAKEN TO BE ZERO)
potential energy = -kqq/d
CHARGES ARE PLACED ON VERTICES OF A SQ. WITH a, b HAS +q & c,d has -q charge . LET E ELECTRIC FIELD & V IS POT. AT CENTRE . IF CHARGES ON a &b R INTERCHANGED WITH THOSE OF c& d RESPTLY THEN E CHANGES WHY??
in this case magnitude will remain same. Only direction of E will change.