Electric potential and field of sphere.

In summary, the total charge in a sphere of radius R is equal to Q. The electric field inside the sphere is given by E = (Q/R^4)*(r^2)/(4i*εo) N/C and outside the sphere it is E = Q/(4*pi*r^2*εo) N/C. The electric potential outside the sphere is V1(r)-V(∞) = Q/(4*pi*r*εo) V. Inside the sphere, the electric potential is V2(r)-V(∞) = Q/(4*pi*R*εo) - ∫(R to r)(Q/R^4)*(r^2)/(4i*εo)dr V.
  • #1
Fabio010
85
0
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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  • #2
Looks good to me. Does the integration limit "a" stand for R in the last couple of lines?
 
  • #3
TSny said:
Looks good to me. Does the integration limit "a" stand for R in the last couple of lines?


yes it does.
 

Related to Electric potential and field of sphere.

1. What is the electric potential of a sphere at a specific point?

The electric potential at a specific point on a sphere is defined as the amount of work required to move a unit positive charge from infinity to that point, against the electric field. It is given by the equation V = kQ/r, where k is the Coulomb's constant, Q is the charge on the sphere, and r is the distance from the center of the sphere to the point.

2. How is the electric potential and field of a sphere related?

The electric field of a sphere is directly proportional to the gradient of its electric potential. This means that the electric field points in the direction of decreasing potential and its magnitude is given by E = -dV/dr, where V is the electric potential and r is the distance from the center of the sphere.

3. What is the equation for the electric field of a charged sphere?

The electric field of a charged sphere is given by the equation E = kQ/r², where k is the Coulomb's constant, Q is the charge on the sphere, and r is the distance from the center of the sphere. This equation applies for points both inside and outside the sphere.

4. What is the significance of the electric potential of a sphere being constant on its surface?

The electric potential of a sphere is constant on its surface because the distance from the center to any point on the surface is the same. This means that the electric potential is the same at all points on the surface. This also means that the work required to move a unit positive charge along the surface of the sphere is zero.

5. How does the electric potential and field change as you move closer or further away from a charged sphere?

As you move closer to a charged sphere, both the electric potential and field increase in magnitude. This is because the distance from the center of the sphere decreases, resulting in a stronger electric field and higher electric potential. As you move further away from the sphere, the electric potential and field decrease in magnitude. This is because the distance from the center of the sphere increases, resulting in a weaker electric field and lower electric potential.

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