Question

Two parallel, infinite, nonconducting plates are $10.0\mathrm{cm}$ apart and have charge distributions of $+1.00\mu \mathrm{C}/{\mathrm{m}}^2$ and $-1.00\mu \mathrm{C}/{\mathrm{m}}^2.$ What is the force on an electron in the space between the plates? What is the force on an electron located outside the two plates near the surface of one of the two plates?

Short answer

Answer: The force on an electron when it is (1) between the parallel plates and (2) outside the plates near the surface of one of the plates is both $9.04\ast {10}^{-15}\mathrm{N}$ towards the positively charged plate.

Step-by-step solution

Determine the electric field between the plates

The electric field between parallel infinite plates can be determined using the charge distribution on the surfaces of the plates. The electric field E equals to: E = σ / (2 * ε₀) where σ is the surface charge density and ε₀ is the vacuum permittivity (approximately $8.85\ast {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$). Now, we plug in the values given in the exercise: σ = $1.00\ast {10}^{-6}\mathrm{C}/{\mathrm{m}}^2$ (for both plates). E = $(1.00\ast {10}^{-6})/(2\ast 8.85\ast {10}^{-12})=5.65\ast {10}^4\mathrm{N}/\mathrm{C}$

Calculate the force on an electron between the plates

Now, we need to find the force on an electron between the plates. We can use Coulomb's Law: F = qE where q is the charge of an electron and E is the electric field found in step 1. The charge of an electron is approximately $-1.6\ast {10}^{-19}\mathrm{C}$. F = $(-1.6\ast {10}^{-19})\ast (5.65\ast {10}^4)=-9.04\ast {10}^{-15}\mathrm{N}$ The force on an electron between the plates is $9.04\ast {10}^{-15}\mathrm{N}$ directed towards the positively charged plate (since the electron is negatively charged).

Determine the electric field outside the plates

The electric field outside the parallel infinite plates equals only the electric field caused by one of the plates because the fields from both plates cancel each other out. In this case, we can use the same formula as in step 1 to find the electric field outside: E_outside = σ / (2 * ε₀) Using the same σ value from the exercise: E_outside = $(1.00\ast {10}^{-6})/(2\ast 8.85\ast {10}^{-12})=5.65\ast {10}^4\mathrm{N}/\mathrm{C}$

Calculate the force on an electron outside the plates

Now let's find the force on an electron when it is outside near one of the two plates, using the same Coulomb's Law from step 2: F_outside = qE_outside Again, the charge of an electron is approximately $-1.6\ast {10}^{-19}\mathrm{C}$. F_outside = $(-1.6\ast {10}^{-19})\ast (5.65\ast {10}^4)=-9.04\ast {10}^{-15}\mathrm{N}$ The force on an electron outside the plates and near the surface of one plate is $9.04\ast {10}^{-15}\mathrm{N}$ directed towards the positively charged plate (since the electron is negatively charged). In conclusion, the force on an electron in the space between the plates is $9.04\ast {10}^{-15}\mathrm{N}$ towards the positively charged plate, and the force on an electron located outside the two plates near the surface of one plate is also $9.04\ast {10}^{-15}\mathrm{N}$ towards the positively charged plate.