A hollow, spherical conducting shell is electrically neutral. Its outer radius I

Question

A hollow, spherical conducting shell is electrically neutral. Its outer radius Is 40 cm. A point charge of 10 C is placed within the cavity. The surface charge density o on the outer surface of the sphere is: The figure shows a non-conducting rod of length L = 1.4 m carrying a non-uniform charge density lambda = alpha.x^2 with alpha = 10^-3 C/m^3. The rod is on the x axis with the left end at the origin. Assuming V = 0 at infinity, find the electric potential (in volts) at the left end of the rod - in other words, at the origin. Select the closest answer. For the same rod and point P of the previous problem, find the electrostatic field at point P.

Explanation / Answer

as no electric field can exist within a conductor, an equal and negative charge will appear on the surface of the conducting shell

hence surface total charge=-10 C

then surface charge density=charge/surface area=-10/(4*pi*radius^2)

=-10/(4*pi*0.4^2)=-4.9736 C/m^2

=-5 C/m^2 approximately

hence option A is correct.

Q10.

potential due to a charge q at a distance of d is given by V=k*q/d

where k=9*10^9 m/F

consider a small strip of charge dx at a distance of x from origin

then charge on this strip=charge density*length

=10^(-3)*x^2*dx

potential at origin (distance of x from this point ) is given by

9*10^9*10^(-3)*x^2*dx/x

=9*10^6*x*dx

integrating this value over x=0 to x=L=1.4 m,
we get

potential at origin=integration of 9*10^6*x*dx

=9*10^6*0.5*x^2

putting limits of x , from x=0 to x=1.4 m

we get

potential at origin=4.5*10^6*(1.4^2-0^2)=8.82*10^6 volts

hence option B is the closest answer.

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