A nonconductive sphere of radius R = 2.00 cm carries a charge Q = 2.00×1012 C un

Question

A nonconductive sphere of radius R = 2.00 cm carries a charge Q = 2.00×1012 C uniformly distributed throughout its volume. (a) Use Gauss’ Law to nd the electric eld inside and outside the sphere, i.e. for the cases r < R and r R. (b) From the electric eld, nd the electric potential V (r) for r < R and for r R, assuming that V = 0 at r = . (c) Evaluate V at r = 1.00 cm.

i think the electric fields will be E=r/(4pi0)*(2.5*10^-11) and E=1/(4pi0)*((2.0*10^-12)/r^2) but do not know where to go from here

Explanation / Answer

Using Gauss law

E*dA = q_inside/eo

draw a gaussian spherical surface passing through a point at r < R

dA = 4*pi*r^2

q_inside = Q*((4/3)*pi*r^3)/((4/3)*pi*R^3)= Q*(r/R)^3

then

E*4*pi*r^2 = (Q/eo)*(r/R)^3

E= (Q/(4*pi*eo))*(r/R^3)

1/(4*pi*eo) = k = 9*10^9 N-m^2/C^2 and Q = 2*10^-12 Cand R = 2 cm

E = (2*10^-12*9*10^9)*r/(0.02)^3 = 2250*r

at r > or = R

Gauss law

E*dA = q_inside/eo

E*4*pi*r^2 = Q/eo

E = (1/(4*pi*eo))*(Q/r^2)

E = k*Q/r^2 = (9*10^9*2*10^-12)/r^2 = 0.018/r^2

b) for r < R,potential is V = integral (E*dr) = integral (2250*r*dr) = 2250*r^2/2 = 1125*r^2

For r> R

V = integeal of (k*Q/r^2)*dr) = -k*Q/r = -9*10^9*2*10^-12/r = -0.018/r

C)

at r = 1 cm ,V = 2250*r^2/2 = 2250*0.01^2/2 = 0.1125 V

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