Question

[img]https://upload.cc/i1/2020/03/13/Iepuz2.jpg[/img]



R is 4

Answer 1

(a) Solution: To find the net charge of sphere A we should multiply the surface charge density by the surface area of this sphere $S_A=4\pi a^2=4\cdot 3.14 \cdot (4.5\cdot10^{-2} m)^2=2.54\cdot 10^{-2} m^2$ $Q_A=78.6 \cdot fC\cdot m^{-2}\cdot S_A=2.00 fC$.

Answer: The net charge of sphere A is $Q_A=2.00\cdot 10^{-12}C$

(b) Solution: The electric field inside uniformly charged spheres is always zero. This can be proved using the Gauss theorem. The electric field in the gap between sphere A and B is determined only by the charge on the inner sphere A. It is Known that the electric field of a spherically distributed charge is identical to the field of a point charge and is calculated using the Coulomb formula.

$E_{Ar}=k\frac{Q_A}{r^2}= 8.99\cdot 10^9[NC^{-2}m^2] \frac{2.00\cdot 10^{-12}C}{(5\cdot 10^{-2}m)^2}=7.18 NC^{-1}$ Here The Coulomb constant is used in Si system $k=8.99\cdot 10^9 NC^{-2}m^2$. As $Q_A>0$ the direction of electric field is radially outward.

Answer: The magnitude of electric field is $E_{Ar}=7.18 NC^{-1}$ with direction outward from the center of A shell.

(c) Solution: The electric field inside uniformly charged spheres B is an sum of field shell A and those charge of shell B which are inner within sphere radius r. It should be assumed that the charge of the shell B is distributed evenly over its volume

(1) $V_B=\frac{4\pi}{3}\cdot (c^3-b^3)$

with density

(2) $\rho_B=\frac{Q_B}{V_B}$ .

The electric charge of shell B at radius $c\ge r\ge b$ will be

(3) $q_{B}(r)=\rho_B \cdot (4\pi/3)\cdot(r^3-b^3)$

and its electric field

$E_{Br}=k\cdot \frac {q_{B}(r)}{r^2}$

Overall electric field

$E_r=E_{Ar}+E_{Br}=k\frac{Q_A}{r^2}+k\cdot \frac {q_B(r)}{r^2}=\frac {k}{r^2}\cdot [Q_A+q_B(r)]$

will be zero if the sum of overall charge within the sphere r will be zero. Thus we find r

$Q_A+q_B(r)=0$ . Substitute (3) (2) (1) in this equation we find

$\\Q_A+\rho_B \cdot (4\pi/3)\cdot(r^3-b^3)=0\\r=(b^3-\frac{Q_A}{\rho_B \cdot (4\pi/3)})^{1/3}=(b^3-\frac{Q_A}{(4\pi/3)Q_B/V_B})^{1/3}=(b^3-\frac{Q_A}{(4\pi/3)Q_B/V_B})^{1/3}$

$r=(b^3-\frac{Q_A(c^3-b^3}{Q_B})^{1/3}=((6.5)^3-\frac{2((7.4)^3-(6.5)^3)}{-5})^{1/3}=6.89cm$ .

Answer: The radial distance r that there will be no electric field in shell B is 6.89 cm

(d) Answer: The electric flux through a concentric Gaussian sphere of radius 9.4 cm will be zero as the sum of charge within it will be $Q_A+Q_B+Q_C=2 -5+3=0$ . The charge of the conducting shell C is distributed over its inner surface with radius d=8.8 cm because it is attracted to the center by the total charge of the shell A and B.