Force on a uniformly charged solid sphere

In summary, Daniel attempted to solve this problem using Maxwell's stress tensor. He first calculated the net force on the northern hemisphere of a uniformly charged sphere of radius R and charge Q. He then used the stress tensor to calculate the force.
  • #1
stunner5000pt
1,461
2

Homework Statement


Griffiths Example 8.2
Determine the net force on the northern hemisphere of a uniformly charged solid sphere of radius R and charge Q. use maxwell's stress tensor to calculate the force. (Not Gauss/Coulomb's law)


Homework Equations


The stress tensor is defined by
[tex] T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{IJ} B^2) [/tex]
where i and j are coordinates

The Attempt at a Solution


ok For the bowl part
[tex] da = R^2 \sin \theta d\theta d\phi \hat{r} [/tex]

[tex] \vec{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{R^2} \hat{r} [/tex]

so
[tex] T_{zx} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \cos\phi [/tex]

[tex] T_{zy} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \sin\phi [/tex]

[tex] T_{zz} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 (\cos^2 \theta - \sin^2 \theta)[/tex]

The net froce is in the Z direction (taking the Z axis to point stragiht up, teh flat part of this bowl lying on the XY plane...

[tex] (T da)_{z} = T_{zx}da_{x}+T_{zy}da_{y}+T_{zz}da_{z} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin\theta \cos\theta d\theta d\phi[/tex]

I don't understand what happened in that last step...
... i know this question seems silly... i should know all of this thoroughly in and out at this point...

anyway why does it turn into that??

what are dax and day and daz??
did they convert into spherical coords and then 'expand' ??

thanks for your help
 
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  • #2
The projections of the [itex] d\vec{a} [/itex] vector onto the 3 cartesian axis.

He made all multiplications and additions and ended up with that result.

Daniel.
 
  • #3
dextercioby said:
The projections of the [itex] d\vec{a} [/itex] vector onto the 3 cartesian axis.

He made all multiplications and additions and ended up with that result.

Daniel.

i didnt see it till i actually looks at da and then i saw it

thanks for your help
 

FAQ: Force on a uniformly charged solid sphere

1. What is the formula for calculating the force on a uniformly charged solid sphere?

The formula for calculating the force on a uniformly charged solid sphere is F = k * (q1 * q2) / r^2, where F is the force in Newtons, k is the Coulomb's constant (9*10^9 Nm^2/C^2), q1 and q2 are the charges in Coulombs, and r is the distance between the two charges in meters.

2. How does the force on a uniformly charged solid sphere change with distance?

The force on a uniformly charged solid sphere is inversely proportional to the square of the distance between the two charges. This means that as the distance increases, the force decreases, and as the distance decreases, the force increases.

3. Does the force on a uniformly charged solid sphere depend on the size of the sphere?

No, the force on a uniformly charged solid sphere does not depend on the size of the sphere. As long as the sphere has a uniform charge distribution, the force will be the same regardless of its size.

4. How does the force on a uniformly charged solid sphere change if the charge is doubled?

If the charge on a uniformly charged solid sphere is doubled, the force will also double. This is because the force is directly proportional to the product of the charges.

5. Can the force on a uniformly charged solid sphere be repulsive?

Yes, the force on a uniformly charged solid sphere can be repulsive if the two charges have the same sign. This is because like charges repel each other, while opposite charges attract each other.

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