# Homework Help: Calculationg the electric field of an insulating sperical shell using integration?

1. Feb 4, 2012

### nosmas

Calculationg the electric field of an insulating sperical shell using integration???

1. The problem statement, all variables and given/known data

We are asked to calculate the electric field at the center of an insulating hemispherical shell with radius R and a uniform surface charge density using integration.

2. Relevant equations
Gauss's law
E*A = Qencl/epsilon

dE = k(dQ/r^2) where k=1/(4pi*epsilon)

3. The attempt at a solution

I assume we can't use gauss's law since it asks for integration so I tried treating it as a ring of charge using dE = k(dQ/r^2)

dQ = lamda * dS
dS = rd(theta)
dQ = lamda * r * d(theta)

dE = k((lamda * d(theta))/r)(cos(theta) i + sin (theta) j)

and integrating from there but i dont believe it can be treated as a ring of charge.

Any suggestions??

2. Feb 4, 2012

### SammyS

Staff Emeritus
Re: Calculationg the electric field of an insulating sperical shell using integration

Hello nosmas. Welcome to PF !

There's not enough symmetry yo use Gauss's Law.

You should start by defining θ.

Shouldn't you be using σ (sigma) for surface charge density ? ... or are you given The total charge, Q, so that σ = Q/(2πR2) ?

dS = r dθ can't be correct. dS has units of area, r dθ has units of length.

If the hemisphere sits on the xy-plane and is centered at the origin, then the x & y components of the E field cancel, leaving only the z-component.

I does make sense to break the hemisphere up into rings of charge.

Assuming θ is the angle of elevation above the xy-plane, then the width of such a ring is Rdθ. The radius of the ring is R cos(θ) . The area of such a ring is the width times the circumference.

That may get you started in the right direction.

3. Feb 6, 2012

### nosmas

Re: Calculationg the electric field of an insulating sperical shell using integration

I should be using σ for the surface charge density I am just not sure how to incorporate it.

So the E field on the hemisphere is equal to the sum of infinetly many rings of charge?

I understand that the width of the ring is Rdθ and the radius of the ring is R cos(θ)

so the area of the ring would be 2∏R(Rdθ) but I dont understand where to use the area?

Using the general equation dE = K*(dQ/r^2) how would i incorporate σ into dQ?

Thanks

4. Feb 6, 2012

### SammyS

Staff Emeritus
Re: Calculationg the electric field of an insulating sperical shell using integration

You left cos(θ) out of your expression for the area of a ring. (I inserted it in the above quote.)

Charge = σ × area → dQ = σ dA = σ (2∏R)cos(θ)(Rdθ)

I don't know how you have the hemisphere oriented in your coordinate system, but ... As I said in my previous post: "If the hemisphere sits on the xy-plane and is centered at the origin, then the x & y components of the E field cancel, leaving only the z-component." In fact this is true for the electric field, dE, due to each ring. So, to find dE, use Coulomb's Law multiplied by sin(θ) .

Integrate the result to find E.