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# The next figure shows, in cross section, a plastic, spherical shell with uniform charge Q = −16e and radius R = 10 cm.?

A particle with charge q = +5e is at the center. We want the electric field (magnitude and direction) at (a) point P1 at radial distance r1 = 6.00 cm and (b) point P2 at radial distance r2 = 12.0 cm.

To find the field at point P1 at radial distance r1, we center a Gaussian sphere on the particle and with P1 on its surface.

Q: in terms of r1, what is the area ∫dA of the Gaussian sphere?

Q2: What, then, is the value of the field magnitude E at radial distance r1= 6.00 cm due to the charged particle?

E= ___ N/C

### 1 Answer

- NCSLv 74 months agoFavorite Answer
Q1)

∫ dA = 4πr₁²

Q2) From your bottom image,

ε₀ * E * ∫dA = q_enc

8.85e-12 C²/N·m² * E * 4π(0.10m)² = 5*1.6e-19C

solves to

E = 7.2e-7 C ◄

Compare that to

E = k*Q / r²

where k = 1/4πε₀ = 8.99e9 N·m²/C²

For r < 10 cm, the outer shell of charge has no effect on the field.

For r > 10 cm, q_enc = -11e = -11 * 1.6e-19C

and you can find the new area, and therefore the new field magnitude, for any r > 10 cm.

Hope this helps!

I dare not edit my original post because a bug on this platform sometimes renders an edited answer "invisible."