51 - Electric Field and Potential Questions Answers
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.
IN THE SOLUTION GIVEN BY AMIT TO MY PREVIOUS QUERY , THE FORMULA USED IS E=kq/r ,SIR I WANT TO KNOW IS IT IS CORRECT , DON'T YOU THINK IT SHOULD BE E=kq/r2 ??
yes you are right
A point charge q is placed off centre at point c inside a thick spherical shelll of inner radius a and outer radius b. Shell is neutral and conducting. OA =ra , OB = rb & OC = rc. Then the potential at centre of shell will be ( also show the distribution of charges )
Distribution: at C q, at inner surface -q, at outer surface +q
so potential at centre = kq/rc -kq/a + kq/b
A particle P of charge q and mass m is placed at a point in gravity free space and its free to move. another particle Q , of same charge and mass , is projected from a distance r from P with an initial speed v0 towards p . Initially the distance between P and Q decreases and then increases. Then what will be the speeds of particles P and Q when their separation is minimum?
by momentum conservation mv0 = 2mv1
so v1 = v0/2