In the figure a proton is a distance d / 2 ?directly above the center of a square of side d. ?What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge d.) ?Number Units”(degrees) ?This answer has no units m kg 5 ?m / ?s m / ?s^? 2 ?N J W N / ?m A V N / ?C or V / ?m C / ?m^? 2 ?C / ?m^? 3 ?N · ?m^? 2 / ?C Units
The Correct Answer and Explanation is :
To determine the electric flux through the square due to a proton located a distance ( d/2 ) above its center, we use Gauss’s Law, which states:
[
\Phi_E = \frac{q_{\text{enc}}}{\epsilon_0}
]
where:
- ( \Phi_E ) is the total electric flux,
- ( q_{\text{enc}} ) is the charge enclosed by a Gaussian surface,
- ( \epsilon_0 ) is the permittivity of free space (( 8.85 \times 10^{-12} ) C²/N·m²).
Step 1: Consider a Cube Enclosing the Proton
The problem hints at treating the square as one face of a cube with edge length ( d ). If we imagine a cube of side ( d ) enclosing the proton, the proton is located symmetrically above the center, implying that its entire charge ( q ) is enclosed by the cube.
Since a cube has six faces, and the charge is enclosed symmetrically, the total electric flux through the cube is:
[
\Phi_{\text{total}} = \frac{q}{\epsilon_0}
]
Since the charge is uniformly distributed among all six faces, the flux through one face (our square) is:
[
\Phi_{\text{square}} = \frac{q}{6\epsilon_0}
]
Step 2: Substitute the Charge of a Proton
A proton has charge:
[
q = 1.6 \times 10^{-19} \text{ C}
]
Thus, the flux through one face is:
[
\Phi_{\text{square}} = \frac{(1.6 \times 10^{-19} \text{ C})}{6 (8.85 \times 10^{-12} \text{ C²/N·m²})}
]
[
\Phi_{\text{square}} \approx 3.01 \times 10^7 \text{ N·m²/C}
]
Answer:
[
\Phi_{\text{square}} \approx 3.01 \times 10^7 \text{ N·m²/C}
]
Explanation:
- We used Gauss’s Law to relate flux to charge.
- We imagined a cube with the square as one of six faces.
- Since the charge is at the center of the cube, flux distributes equally among all faces.
- We divided the total flux by 6 to get the flux through one face.
- Finally, we substituted the charge of a proton to compute the numeric result.
Thus, the electric flux through the square is ( 3.01 \times 10^7 ) N·m²/C.