Electric Field inside and outside of Dielectric sphere

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SUMMARY

The discussion focuses on determining the electric field inside and outside a dielectric sphere of radius R with a uniform permanent polarization P. The solution involves applying Laplace's equation to the electrostatic potential, represented as V, and utilizing the relationship D = ε₀E + P. The boundary conditions dictate that outside the sphere, the potential approaches zero as r approaches infinity, while inside the sphere, the potential must remain finite as r approaches zero. The coefficients Al and Bl need to be calculated to fully define the electric field in both regions.

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  • Understanding of Laplace's equation in electrostatics
  • Familiarity with electric displacement field D and polarization P
  • Knowledge of boundary conditions in electrostatic problems
  • Basic concepts of dielectric materials and their properties
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Homework Statement



A dielectric sphere of radius R has a uniform permanent polarisation P. Determine the electric field both inside and outside the sphere. (Hint: Since div P = 0 everywhere, the electrostatic potential satisfies Laplace's equation. Do not assume that the sphere is characterised by a dielectric constant. Instead, use D = \epsilon_{0}E0 + P.)

Homework Equations



use the laplace equation where the general form of potential is:
V=\Sigma (A_{l}r^{l}P_{l}(cos(theta))+B_{l}*1/(r_{l+1})*P_{l}(cos(theta)))

The Attempt at a Solution



set the boundary condition when outside of sphere :
r ->infinity, V->0 hence Al=0

when inside the sphere
r-> 0, V not= infinity hence Bl=0

the next time is to find both the coefficients, a.k.a Al and Bl. so far am i heading the right directions?my question is also what D = \epsilon_{0}E0 + P purpose here in solving this problem?
 
Last edited:
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There are two regions in space, one inside and one outside the sphere. Therefore, you need to consider two separate potentials, one inside and one outside, then match boundary conditions. That's where the equation ##\mathbf{D}=\epsilon_0 \mathbf{E}+\mathbf{P}## comes in.
 

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