Respuesta :
Answer:
(a) [tex]-6.76\times 10^{12}\ N/C[/tex]
(b) [tex]-1.352\times 10^{13}\ N/C[/tex]
(c) [tex]-7.2\times 10^{11}\ N/C[/tex]
Explanation:
(a)
Given:
Charge on sphere (Q) = [tex]-20\ \mu C=-20\times 10^{-6}\ C[/tex]
Radius of sphere (R) = 11.0 cm = 0.110 m
Distance from the center (r) = 5 cm = 0.05 m
Coulomb's constant (k) = [tex]9\times 10^{9}\ Nm^2/C^2[/tex]
Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:
[tex]E=(\frac{kQ}{R^3})r[/tex]
Plug in the given values and solve for 'E'. This gives,
[tex]E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)[/tex]
(b)
Given:
Charge on sphere (Q) = [tex]-20\ \mu C=-20\times 10^{-6}\ C[/tex]
Radius of sphere (R) = 11.0 cm = 0.110 m
Distance from the center (r) = 10 cm = 0.10 m
Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:
[tex]E=(\frac{kQ}{R^3})r[/tex]
Plug in the given values and solve for 'E'. This gives,
[tex]E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)[/tex]
(c)
Given:
Charge on sphere (Q) = [tex]-20\ \mu C=-20\times 10^{-6}\ C[/tex]
Radius of sphere (R) = 11.0 cm = 0.110 m
Distance from the center (r) = 50 cm = 0.50 m
Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:
[tex]E=\dfrac{kQ}{r^2}[/tex]
Plug in the given values and solve for 'E'. This gives,
[tex]E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)[/tex]
