Question

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Answer 1

The logic and analysis of the solution above is absolutely right, i just find there to be some errors in the setting up of the integral and the following solution. Here is what i get when i solve the problem.

Flux = Integral (E.dA) = Qenclosed/Eo

For the Integral(E.dA) part of the equation for flux we get that,
Integral(E.dA)=E*2*pi*R*L

For the Qenclosed/Eo part of the equation for flux we get that,
Qenclosed/Eo = Integral(?.dV)/Eo

We are given that ? = A*r2 and we know that ? = dq/dv. Therefore we get that
Qenclosed = Integral(?.dV)..... V = pi*r^2*L; dV = 2*pi*r*L.dr

Qenclosed = Integral(?.2*pi*r*L.dr) = 2*pi*L*Integral(?.r.dr)

so we have
Qenclosed = 2*pi*r*L*A*Indergral(r^3.dr)

So far on the left hand side we have;
flux = E*2*pi*R*L
and on the right hand side we should have;
flux = (2*pi*r*L*A*r^4)/4Eo

equating the both the equation for Electric field should be

E = A*r^3/(4*Eo) where R = r = 6 cm

Part b) Now since the gaussian cylinder is outside our real cylinder (r = 7 cm), we would use the same equation as above with the only difference that R = 8 cm & r = 7cm. So the equation would be

E = A*r^4/(4*Eo*R) where R = 8 cm & r = 7 cm...