Calculating Electric Field of Insulating Spherical Shell

[nosmas]

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

Homework Statement

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.

Homework Equations

Gauss's law
E*A = Qencl/epsilon

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

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 don't believe it can be treated as a ring of charge.

Any suggestions??

 

[SammyS]

Homework Statement

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.

Homework Equations

Gauss's law
E*A = Qencl/epsilon

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

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 don't believe it can be treated as a ring of charge.

Any suggestions??

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.

 

[nosmas]

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 don't understand where to use the area?

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

Thanks

 

[SammyS]

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 infinitely 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(cos(θ))(Rdθ) but I don't understand where to use the area?

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

Thanks

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.