Respuesta :
Answer:
The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C
Explanation:
Step 1: Data given
Charge of uniform density of first sphere= 40 pC/m² = 40.0 * 10^-12 C/m²
Radius of first sphere is 1.0 cm
Radius of second sphere = 3.0 cm
Charge of uniform density of second sphere= 60 pC/m² = 60.0 * 10^-12 C/m²
Step 2: Calculate the magnitude
Sphere surface area = 4πr²
Charge on inner sphere Qi = 40.0*10^-12C/m² * 4π(0.01m)² = 5.027*10^-14 C
NET charge on outer sphere Qo = 60.0*10^-12 * 4π(0.03m)² = 6.786*10^-13 C
Inner sphere induces a - 5.027*10^-14 C charge (-Qi) on inside of the outer shell
This means there is a net zero charge within the outer shell.
For the outer shell to show a NET charge +6.786*10^-13C, it's must have a positivie charge {+6.786*10^-13C + (+5.027*10^-14C)} = +7.2887*10^-13 C
Regarding the outer shell as a point charge (field at 0.04m is)
E = kQ /r²
E = (8.99^9)*(7.2887*10^-13 C) / (0.04)² .. .. ►E = 4.10 N/C
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Using Gauss's law .. with a spherical Gaussian surface at 0.04m enclosing a net charge +7.2887*10^-13 C
Flux EA = Q / εₒ
⇒ with εₒ = 8.85 *10^-12
⇒ with Q = 7.2887*10^-13 C
E * 4π(0.04)² = 7.2887*10^-13 C / 8.85*10^-12
E = 7.2887*10^-13 C / {8.85*10^-12 * 4π(0.04)²}
E = 4.10 N/C
The magnitude of the electric field at a point 4.0 cm from the center of the two surfaces is 4.10 N/C
A correct option is option (2).
Given,
The Surface area of a sphere [tex]A=4\pi r^2[/tex]
Find the charge on the inner sphere.
[tex]Q_i =40.0^{2} c/m^2\times4\pi (0.01m)^2\\=5.03^{-14 C}[/tex]
In this case, the Inner sphere induces a [tex]5.03^{-14 C}[/tex]charge [tex](-Q_i)[/tex] on the inside of the outer shell.
Therefore, there is a net-zero charge within the outer shell.
For the outer shell to show a net charge [tex]+6.79^{-13}C[/tex], it's must have a +ve charge.
[tex]{+6.79^{-13} C + (+5.03^{-14} C)} = +7.29^{-13} C[/tex]
Now, Use Gauss's law, with a spherical Gaussian surface at 0.03m enclosing a net charge[tex]+7.29^{-13} C[/tex]
[tex]Flux=E_A\\=\frac{Q}{\varepsilon _0 }[/tex]
Substitute numerical values we get,
[tex]E\times4\pi (0.04)^2=\frac{.29^{-13} }{8.85^{-12} } \\E=4.5 N/C[/tex]
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