Question
Four dipoles, each consisting of a +10 µC charge and a -10 µC charge, are located in the xy-plane with their centers 1.0 mm from the origin, as shown. A sphere passes through the dipoles, as shown in the figure. What is the electric flux through the sphere due to these dipoles? (ε0 = 8.85 × 10-12C2/N m2)
The sphere only encloses the positive charge in each dipole.
Qu1-pic1-2020
9.0 × 106 N m2/C
0.00 N m2/C
11 × 105 N m2/C
4.5 × 106 N m2/C
The sphere only encloses the positive charge in each dipole.
Qu1-pic1-2020
9.0 × 106 N m2/C
0.00 N m2/C
11 × 105 N m2/C
4.5 × 106 N m2/C
Answers
GPT-4o mini
To determine the electric flux through the sphere due to the dipoles, we can use Gauss's law, which states that the electric flux \(\Phi_E\) through a closed surface is related to the charge enclosed \(Q_{\text{enc}}\) within that surface by the equation:
\[
\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}
\]
Given that the sphere encloses only the positive charge of the dipoles, we first need to calculate the total charge enclosed by the sphere. Each dipole consists of a +10 µC charge and a -10 µC charge. Since the sphere only encloses the positive charge in each dipole and there are four dipoles, the total enclosed charge is:
\[
Q_{\text{enc}} = 4 \times (+10 \, \mu C) = 4 \times 10 \times 10^{-6} \, C = 4 \times 10^{-5} \, C
\]
Next, we substitute this value of \(Q_{\text{enc}}\) into Gauss's law. The permittivity of free space \(\varepsilon_0\) is provided as \(8.85 \times 10^{-12} \, C^2/(N \cdot m^2)\). Thus, we can calculate the electric flux:
\[
\Phi_E = \frac{4 \times 10^{-5} \, C}{8.85 \times 10^{-12} \, C^2/(N \cdot m^2)}
\]
Calculating the above expression,
\[
\Phi_E = \frac{4 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 4.52 \times 10^6 \, N \cdot m^2/C
\]
To provide the final answer regarding the electric flux through the sphere due to the dipoles:
\[
\Phi_E \approx 4.5 \times 10^6 \, N \cdot m^2/C
\]
Therefore, the correct answer is **4.5 × 10^6 N · m²/C**.
\[
\Phi_E = \frac{Q_{\text{enc}}}{\varepsilon_0}
\]
Given that the sphere encloses only the positive charge of the dipoles, we first need to calculate the total charge enclosed by the sphere. Each dipole consists of a +10 µC charge and a -10 µC charge. Since the sphere only encloses the positive charge in each dipole and there are four dipoles, the total enclosed charge is:
\[
Q_{\text{enc}} = 4 \times (+10 \, \mu C) = 4 \times 10 \times 10^{-6} \, C = 4 \times 10^{-5} \, C
\]
Next, we substitute this value of \(Q_{\text{enc}}\) into Gauss's law. The permittivity of free space \(\varepsilon_0\) is provided as \(8.85 \times 10^{-12} \, C^2/(N \cdot m^2)\). Thus, we can calculate the electric flux:
\[
\Phi_E = \frac{4 \times 10^{-5} \, C}{8.85 \times 10^{-12} \, C^2/(N \cdot m^2)}
\]
Calculating the above expression,
\[
\Phi_E = \frac{4 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 4.52 \times 10^6 \, N \cdot m^2/C
\]
To provide the final answer regarding the electric flux through the sphere due to the dipoles:
\[
\Phi_E \approx 4.5 \times 10^6 \, N \cdot m^2/C
\]
Therefore, the correct answer is **4.5 × 10^6 N · m²/C**.