A point charge 2q is placed at the origin (x=0, y=0). The center of a thick cond
ID: 1957837 • Letter: A
Question
A point charge 2q is placed at the origin (x=0, y=0). The center of a thick conducting spherical shell carrying a net charge of -6q is also at the origin (x=0, y=0). The conducting shell has an inner radius R1 and outer radius R2.a) what is the electric field at a point a distance r from the origin, where 0<r<r1?
b) what is the electric field at a point a distance r from the origin, where R1<r<R2?
c) what is the electric field at a point a distance r from the origin, where r>R2?
d) what is the net charge on the outer surface of the spherical shell?
Explanation / Answer
a.) Electric field inside the shell cavity (0
Inside a charged shell, field due to the shell itself is necessarily 0. (as per gauss law, only enclosed charge counts)
Thus the only field will be due to the point charge 2q, which u can calculate by standard couloumb's law.
(alternatively, take gaussian surface of radius 'r'<R1 --> charge enclosed = 2q --> flux() = 2q/0 --> E = /Area
E = 2q/4r20; Which is the same as the couloumbic expression)
b.) Electric field inside the material of the shell (R1<r<R2)
Field inside a conducting material is always 0. (This will help in solving part d also)
c.) Field outside the shell (r>R2)
On calculating the field out side, a shell behaves like a point charge concetrated at its centre. thus, use net charge = 2q + (-6q) = -4q and use couloumb's law again.
(alternatively, take gaussian surface of radius r>R2, net charge enclosed = -4q, procced as step (a.))
d.) since field inside the conductor is 0, any gaussian surface with R1<r<R2 must enclose 0 net charge. thus, the charge on the inside surface of the shell must negate the point charge
thus charge on the inner surface = -2q
since net charge on the shell is -6q
charge on the outer surface =-4q.
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