The electric field on the surface of a 10 -cm-diameter sphere is perpendicular to the sphere and has magnitude 47 kN/C. What's the electric flux through the sphere?

Gauss's law is a foundational principle in electromagnetism that relates the distribution of electric charge to the resulting electric field. It states that the electric flux through a closed surface, also known as a Gaussian surface, is equal to the net charge within the surface divided by the permittivity of the medium. Mathematically, it is expressed as \( \Phi_E = \frac{Q_{enc}}{\epsilon_0} \) where \( \Phi_E \) is the electric flux, \( Q_{enc} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space.

In simpler terms, Gauss's law helps us understand how charge can create an electric field and, conversely, how the field can tell us something about the charge producing it. For a closed surface like a sphere, when the electric field is uniform and perpendicular to the surface, the flux can be easily computed as the product of the field strength and the area of the surface—as is the case in the provided exercise.

The electric field is a vector field that surrounds electric charges and exerts a force on other charges within the field. The strength or magnitude of this field is measured in Newtons per Coulomb (N/C) and is defined as the force experienced by a positive unit charge placed in the field. Electric fields are represented by electric field lines, with the direction of the field being tangent to the lines at any point and the density of the lines indicating the field's strength.

In our current example, the presence of a uniform electric field of 47 kN/C implies that a positive charge placed on the surface of the sphere would experience a force of 47 kN acting perpendicular to the surface. Understanding the direction and magnitude of electric fields is crucial in calculating the electric flux, especially when applying Gauss's law.

The surface area of a sphere is a measure of the total area that the surface of the sphere occupies. For any sphere, this surface area can be found using the formula \( A = 4\pi r^2 \) where \( r \) is the radius of the sphere. When performing calculations related to electric flux and Gauss's law, knowing the surface area is key as it directly impacts the total flux through the surface.

To calculate the correct surface area in our exercise, we must make sure to use the radius, not the diameter. The provided diameter of 10 cm (which is the total distance across the sphere through the center) should be halved to get the radius of 5 cm. This is a subtle yet critical step to ensure the accuracy of our electric flux calculation.