Knowee
Questions
Features
Study Tools

The electric field just above the face of a penny is 1.14×103 N/C. What is the surface charge density on the surface of that penny face?

Question

The electric field just above the face of a penny is 1.14×103 N/C. What is the surface charge density on the surface of that penny face?

🧐 Not the exact question you are looking for?Go ask a question

Solution

The surface charge density (σ) on a surface can be calculated using Gauss's law. According to Gauss's law, the electric field (E) near a charged surface is given by the equation:

E = σ / ε0

where ε0 is the permittivity of free space, which is a constant equal to 8.85 × 10^-12 C^2/N·m^2.

We can rearrange this equation to solve for σ:

σ = E * ε0

Substituting the given values:

σ = (1.14 × 10^3 N/C) * (8.85 × 10^-12 C^2/N·m^2)

σ = 1.0086 × 10^-8 C/m^2

So, the surface charge density on the surface of the penny face is approximately 1.0086 × 10^-8 C/m^2.

This problem has been solved

Similar Questions

Based on the density of your penny, which metal is your penny mostly made of? Prove it using the densities below.Zinc (Zn) = 7.1 g/ml Silver (Ag) = 10.5 g/ml Gold (Au) = 19.3 g/mlCopper (Cu) = 8.9 g/ml Platinum (Pt) = 20.5 g/ml Brass (alloy) = 8.40 g/ml

A surface charge density of 2.10×10-9 C/cm2 exists on a circle-shaped disc. If the disc's radius increases by a factor of 2 (but the total charge stays the same), what is the surface charge density?

A -2.34-nC point charge is located at the center of a conducting spherical shell. The shell has an inner radius of 2.00 m, an outer radius of 4.00 m, and a charge of 8.53-nC. (a) What is the magnitude of the electric field at r = 1.00 m?  N/C (b) What is the magnitude of the electric field at r = 3.00 m?  N/C (c) What is the magnitude of the electric field at r = 5.00 m?  N/C (d) What is the surface charge density, σ, on the outside surface of the shell? (You may enter your calculation using scientific notation.)  C/m2C/m2

A 14 cm-radius hemisphere contains a total charge of853.1 10 C− . The flux through therounded portion of the surface is4 211 10 /Nm C . What is the flux through the flat baseof the hemisphere?

A uniform linear charge density of 4.0 nC/m is distributed along the entire x axis. Consider a spherical (radius = 5.0 cm) surface centered on the origin.Determine the electric flux through this surface.Select one:68 N m2/C62 N m2/C45 N m2/C79 N m2/C23 N m2/C

1/2

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.