(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A spherical charged ball of radius "a" has an empty spherical cavity, of radius b<a, at its center. There is no charge outside the ball and no sheet-charge on its outer surface. The (radial) field has given value Ea on the outer surface; inside the ball it is given as

E(r)= 0 for 0≤r<b

E(r)= (Ear^2)/a^2

a) Evaluate the total charge by using the integral form of Gauss' law.

b) Evaluate the volume charge density ρ(r) for 0<r<a from the differential form of Gauss's Law.

c) Integrate your answer to part b to evaluate the total volume charge (between a and b)

d) Specify the radius of any sheet-charge and evaluate the surface-charge density on it, directly from the given information.

e) Show that your answers to parts c and d agree with your answer to part a.

f) Evaluate E(r) outside the ball, r>a, and complete the graph of this function.

g) Evaluate the potential function at any r greater than or equal to zero, and graph this function.

2. Relevant equations

σ=Q/A

A=4π^2

∫E.ndA=(1/εo)∫pdV

∇.E=ρ/εo

Q=∫pdV

Electric flux= EA=Q/εo

Potential V(x)=-∫E(r)dr+C

3. The attempt at a solution

a) Since there is no sheet charge on its outer surface or charge outside the ball I assume I can place gaussian surface anywhere

Electric flux= EA=Q/εo

EdA=Q/εo

(Ea*r^2)/a^2)(4πr^2)=Q/εo

Q=(4πε0Ea*r^4)/(a^2)

but the answer is 4πεoEa*a^2

My guessed would be r=a since the gaussian surface area could be anywhere, then

Q=(4πε0Ea*r^4)/(a^2)

Q=(4πε0Ea*a^4)/(a^2)

Q=4πε0Ea*a^2 ?

b) ∇.E=ρ/εo

ρ=∇.E*εo

ρ=(εo)[(1/r^2)(d/dr{r^2E(r)})]

ρ=(εo)[(1/r^2)(d/dr{r^2*Ea*r^2)/a^2})]

ρ=(εo)[(1/r^2)(d/dr{r^4*Ea)/a^2})]

ρ=(εo)[(1/r^2)(4r^3*Ea)/a^2)]

ρ=(εo)(4r*Ea)/a^2

ρ=(4rεo*Ea)/a^2 ?

c)Q=∫ρ(r)dV

Q=from b to a∫[(4rεo*Ea)/a^2](4π^2dr)

Q=∫(16πr^3εo*Ea)/a^2dr

Q=(16πεo*Ea)/a^2∫r^3dr

Q=(16πεo*Ea)/a^2 [r^4/4] a to b

Q=(4πεo*Ea)/a^2 [r^4] a to b

Q=[(4πεo*Ea)/a^2](a^4-b^4) ?

d)&e) I'm lost

f) I think this question is too tricky because it seems so simple. If we set the gaussian surface outside the sphere the Electric flux will equal EA, just like the first question a) except that this time we don't use (Ea*r^2)/a^2). So

Electric flux= EA=Q/εo

E(r)=Q/εo*A

E(r)=Q/(4πr^2*εo) ?

g) V(x)=-∫E(r)dr, now 0≤r<∞

V(x)= -∫E(r)inside dr + ∫E(r) outside dr

V(x)= -∫Ea*r^2)/a^2 dr + ∫Q/(4πr^2*εo) dr

V(x)=-(Ea/a^2 (from a to r)∫r^2 dr + Q/4πεo (from ∞ to r)∫1/r^2 dr )

V(x)=- Ea/a^2 [r^3/3]from a to r + Q/4πεo [1/r] from ∞ to r

V(x)=- Ea/3a^2 [r^3-a^3] + Q/4πεo [1/r-0]

V(x)= Q/4πrεo - Ea/3a^2 [r^3-a^3]

V(x)= Q/4πrεo - Ea*r^3/3a^2 + Ea*a^3/3a^2

V(x)= (3Qa^3-QEa*r^3+QEa*a^3)/4πrεo3a^2

V(x)= Q(a^3-Ea*r^3+Ea*a^3)/12πrεoa^2 ?

I'm willing to go letter by letter step by step

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: E&M Charged disk

**Physics Forums | Science Articles, Homework Help, Discussion**