Question

The electric potential between two parallel plates is given by V(x)=(8.0 V/m)x+5.0 V, with x=0 What is the charge density on the plates? taken at one of the plates and x positive in the direction toward the other plate. Express your answer using two significant figures.

Answer 1

The charge density on the plates is -9.0 x 10^-11 C/m².

Step-by-step explanation:

The charge density on the plates can be found by taking the derivative of the electric potential equation with respect to x. In this case, the electric potential equation is V(x) = (8.0 V/m)x + 5.0 V. Taking the derivative, we get dV(x)/dx = 8.0 V/m. The charge density, denoted as σ, is equal to the negative of the derivative of the electric potential, divided by the electric field constant ε₀. Therefore, the charge density on the plates is σ = -8.0 V/m / ε₀.

Using the value of ε₀, which is approximately 8.85 x 10^-12 C²/Nm², we can calculate the charge density. σ = -8.0 V/m / (8.85 x 10^-12 C²/Nm²) = -9.04 x 10^-11 C/m². Finally, expressing the charge density using two significant figures, the charge density on the plates is -9.0 x 10^-11 C/m².