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
- NCSLv 77 months agoFavorite 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
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!