Question

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





R is 4

Answer 1

As per the given question,

Radius of the inner sphere $a=4.5 cm=4.5\times 10^{-2}m$

Surface charge density on the sphere $(\rho_a)= 78.6fc/m^2$

Shell B is made of non conducting material,

inner radius of B $(b)=6.5 cm =6.5\times 10^{-2}m$

outer radius of B, $(c)=(R+70)mm$

a)

As A is conducting sphere, so all the charges given to the sphere A will come to the surface of the sphere A,

Hence total charge $=4\pi a^2\times \rho_a=78.6\times 10^{-15}\times4\pi\times (4.5\times 10^{-2})^2=19.99\times 10^{-16}\simeq2 fC$

b) electric field at a distance r=5 cm

$E=\dfrac{q}{4\pi \epsilon_o R^2}$

$E=\dfrac{2\times 10^{-15}\times 10^9\times 9}{5\times 5\times 10^{-4}}=0.72\times 10^{-2}=7.2\times 10^{-3}N/C$

C)Charges will induce on the surface of the shell B, in the inner surface of the shell B, charges will be -negative due to the the induction and other of the shell B, will be positively charged.

So, net field in the shell B will be zero.

charge on the inner surface =-$-3fC$

Charge on the outer surface of the sphere = 3fc

$\dfrac{-2\times 10^{-15}}{4\pi\times \epsilon_o\times (a+x)^2}+\dfrac{2\times10^{-15}}{4\pi\epsilon_o(c-x)^2}$

everything will get cancel, only x terms will remain in the above equation,

$x=\dfrac{c-a}{2}=\dfrac{8.8-4.5}{2}=\dfrac{4.3}{2}=2.15cm$

d) $Q_{net}=2-5+3=0$

So, from the gaussian rule,

$E=\dfrac{Q_{net}}{\epsilon_o}=\dfrac{0}{\epsilon_o}=0$