Consider the following three concentric systems: two thick shells and a solid sp

Question

Consider the following three concentric systems: two thick shells and a solid

sphere (all conductors, shown in dark colors). The radii, in the increasing order,

are a,b,c,d, and e. The sphere is electrically neutral. The smaller shell is given an

excess charge of -5 C and the larger shell is given an excess charge of +7 C. The

system quickly comes to electrostatic equilibrium.

(a) What are the electric charges (values and signs) on the each of the 5 conducting

surfaces? Are these charges distributed uniformly?

NAME:

(b) Use the Gauss%u2019s law to find expressions (just formulas)for the magnitude of the E-fields in all 6 regions (e.g. 0 < r < a, a < r < b , etc.). Make sure to indicate the directions of the fields as well. (NOTE: You may simply write down the answer just by identifying the %u201Cenclosed charges%u201D. )

Consider the following three concentric systems: two thick shells and a solid sphere (all conductors, shown in dark colors). The radii, in the increasing order, are a,b,c,d, and e. The sphere is electrically neutral. The smaller shell is given an excess charge of -5 C and the larger shell is given an excess charge of +7 C. The system quickly comes to electrostatic equilibrium. What are the electric charges (values and signs) on the each of the 5 conducting surfaces? Are these charges distributed uniformly? Use the Gauss%u2019s law to find expressions (just formulas)for the magnitude of the E-fields in all 6 regions (e.g. 0

Explanation / Answer

There cannot be any electric field in the shells and inside the sphere at equilibrium i.e. r<a, b<r<c, and d<r<e.

shell 1 : d<r<e

assume a imaginary sphere with radius r in this region, the net charge contained in this sphere should be ZERO (gauss's flux law)

suppose charge on outer surface (r=e) is q1, so the charge on inside surface (r=d) is (7-q1)

then the net charge inside the assumed sphere would be

q_net = (7-q1) + (-5) +0.

E=0 ==> q_net =0 ==> q1 = 2 C.

hence the charge on inside surface of outer shell (r=d) is 5 C.

similarly assume the charge on surface r=c is q2. hence the charge on inside surface (r=b) is (-5-q2)

again applying gauss law at region c<b<r,

q_net=0

==> -5-q2 =0

q2 = -5 C.

and there will be no charge on surface r=a, since the condition of zero electric field is already satisfied.

hence the charge distribution will be

   r           a b   c     d     e

   q          +2    +5    -5     0     0

yes the charges will be distributed uniformly, since the system is having radial symmetry

part b:

apply the gauss law in each of the five region, like this

                surface_integral (E)   = q_enclosed / (epsilon_0)

due to radial symmetry E will be constant at a given radius, hence

                      ==> E*4pi*r^2   = q_enclosed / (epsilon_0)

                      ==> E   = q_enclosed / (epsilon_0 * 4pi*r^2)

the direction of the field will be radially outward.

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