Electric potential and field of sphere.

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SUMMARY

The discussion focuses on the electric potential and electric field of a charged sphere with a radius R and a charge density defined by p(r) = Q*r/(pi*R^4). The total charge within the sphere is confirmed to equal Q through integration of the charge density. The electric field inside the sphere (r < R) is derived as E = (Q/R^4)*(r^2)/(4ε₀) N/C, while outside the sphere (r > R), it is E = Q/(4πr²ε₀) N/C. The electric potential is calculated for both inside and outside the sphere, with the potential outside given by V(r) = Q/(4πrε₀) and the potential inside derived through integration of the electric field.

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Fabio010
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In a sphere of radius R the charge density is given by:

p(r) = Q*r/(pi*R^4) , the r is the distance of a generic point to the center of sphere.a) Confirm that the total charge is equal to Q.

b) What is the electric field inside and outside sphere.

c) What is the electric potential inside and outside sphere when the zero potential is considered in infinite and the potential function is continuous in r=R.Attempts:

a) dq = ∫p dV = ∫Q*r/(pi*R^4) dV= Q(pi*R^4)/∫rdV

in spherical coordinates we obtain:

Q/(pi*R^4)∫(0->2pi) ∫(0->pi)∫(0->r) r*r^2*sinθdrd[itex]\Phi[/itex]

Q/(pi*R^4)*(4pi*r^4/4pi)

to have the total charge in sphere r must be = R so

Q/(pi*R^4)*(4pi*R^4/4) = Qb) Inside sphere r < R

so the inner charge is = (Q/R^4)*(r^4)

then by the electric flux

E(4*pi*r^2) = (Q/R^4)*(r^4)*(1/εo) vector E have radial direction

E = (Q/R^4)*(r^2)/(4i*εo) N/COutside sphere r>R

the charge is Q

then by the electric flux

E(4*pi*r^2) = Q/εo

E = Q/(4*pi*r^2*εo) N/Cc) The electric potential outside the sphere is :

r>R

Point 1 = arbitrary point outside sphere
V1(r)-V(∞) = -∫(infinity to r) Q/(4*pi*r^2*εo) dr = Q/(4*pi*r*εo) Vattempt 1:

Point 2 = arbitrary point inside sphere
V2(r)-V(∞) = -∫(infinity to R) Edr (r>R) - ∫(R to r)Edr (r<R) =
= Q/(4*pi*R*εo) - ∫(R to r)(Q/R^4)*(r^2)/(4i*εo)dr ...Can somebody check if the solutions in a and b are correct?. And c, am i proceeding it correctly?
 
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Looks good to me. Does the integration limit "a" stand for R in the last couple of lines?
 
TSny said:
Looks good to me. Does the integration limit "a" stand for R in the last couple of lines?


yes it does.
 

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