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An insulating sphere of radius a, centered at the origin, has a uniform volume charge density p.?

A spherical cavity is excised from the inside of the sphere. The cavity has radius a/4 and is centered at position h, where |h| < 3/4, so that the entire cavity is contained within the larger sphere. Find the electric field inside the cavity.

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The Best asnwer for An insulating sphere of radius a, centered at the origin, has a uniform volume charge density p.?

Throughout this E, r, h, and d will denote vectors.

First find the electric field a a sphere without the cavity:

q = ρV = ρ(4/3*πr^3)

E = kq/r^2 = kρ(4/3πr^3)/r^2 = 4kρπr/3 = ρr/(3ε_0)

A cavity has no charge, which is the same as having two equal and opposite charges. Find the electric field of such an opposite sphere:

E_opp = -ρd/(3ε_0)

This is the same as above, just using d to denote the distance from the center of the opposite sphere. Because the sphere is centered at h in the sphere, d = r - h:

E_opp = -ρ(r - h)/(3ε_0)

E_cav = E + E_opp = ρr/(3ε_0) - ρ(r - h)/(3ε_0) = ρh/(3ε_0)

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************word: AI P, I basically crunched the numbers. the two procedures, yours and mine are perfect. We did an identical element basically in diverse steps. You misinterpret me, and that i misinterpret you. I used the whole charge latest in (a) to calculate the E field, and to that end I had a fraction of entire charge (no longer of density). You used the charge DENSITY given, and greater advantageous by using the quantity of the indoors sphere. the two procedures provide an identical answer. No annoying emotions =). And wow, that replaced into an severe conflict of physics =). *********************************** i visit describe a thank you to physique of innovations those issues, yet i'm no longer likely to clean up them right here for you. It you earnings you greater in case you artwork by using those on your very own =). *******First be useful to transform all gadgets to meters and Coulombs. (a) the whole charge of the sector is the quantity charge density situations the quantity of the sector: charge of sphere = (volume) * (rho) the quantity of a sphere is 4/3 * pi * r^3. you're given rho and r. Convert gadgets and use the above equation. (b) First we word that 2cm is interior the sector. So the only charge that we are in contact approximately is the fraction of the whole charge it fairly is contained interior of a 2cm radius: So Qint = (volume with r = 2cm) / (volume with r = 7cm) * Q_total = 2^3 / 7 ^3 * Q_total = 0.023 * Q_total The E field of a sphere acts like the E field of a ingredient charge with an identical charge. So E = one million /( 4*pi* e0) * Qint * one million/( r^2 ) use r = 2cm , Qint = .023 * Q_total (use Q from section (a) for Q_total) , and one million/(4*pi* e0) = 9 * 10^9 be useful to transform all gadgets. (c) it is comparable to (b), different than that this time 11cm is exterior of the sector. So the E field would be comparable to a ingredient charge. E = one million /( 4*pi* e0) * Q_total * one million/( r^2 ) the place r = 11 cm and Q_total is Q latest in (a) be useful to transform all gadgets first. i wish this facilitates

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