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M asked in Science & MathematicsPhysics · 7 months ago

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

Attachment image

1 Answer

  • NCS
    Lv 7
    7 months ago
    Favorite Answer


    ∫ 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!

    • ...Show all comments
    • NCS
      Lv 7
      7 months agoReport

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

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