The charge so extremely large nonconducting horizontal sheets each carry uniform charge density...

Question:

The charge so extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upper sheet carries +5.00 C/m{eq}^2 {/eq}. The electric field midway between the sheets is {eq}4.25 \times 10^5 {/eq} N/C pointing downward. What is the surface charge density on the lower sheet?

Electric Field due to Sheet of Charge

The electric field due to a large uniformly charged plane depends upon the surface charge density of the plane and the electrical permittivity of the medium around the plane, Mathematically

{eq}\begin{align} E = \frac{\sigma}{2 \epsilon_0} \end{align} {/eq}

Where {eq}\sigma {/eq} is the surface charge density and {eq}\epsilon_0 = 8.85 \times 10^{-12} \ \rm F/m {/eq} is the permittivity of the free space.

Answer and Explanation:

Data Given

  • Surface charge density of the upper sheet {eq}\sigma_1 = 5.00 \ \rm C/m^2 {/eq}
  • The electric field midway between the sheets {eq}E = 4.25 \times 10^5 \ \rm N/C {/eq}

As the electric field is downwards so the other sheet i.e. the lower sheet is negatively charged so we know that electric field due to a sheet of charge is

{eq}\begin{align} E = \frac{\sigma_1}{2 \epsilon_0} + \frac{\sigma_2}{2 \epsilon_0} \end{align} {/eq}

{eq}\begin{align} \frac{\sigma_2}{2 \epsilon_0} = E - \frac{\sigma_1}{2 \epsilon_0} \end{align} {/eq}

{eq}\begin{align} \frac{\sigma_2}{2 \epsilon_0} = 4.25 \times 10^5 \ \rm N/C - \frac{5.00 \ \rm C/m^2}{2 \times 8.85 \times 10^{-12} \ \rm F/m} \end{align} {/eq}

{eq}\begin{align} \frac{\sigma_2}{2 \epsilon_0} = -5.65 \times 10^{11} \ \rm N/C \end{align} {/eq}

{eq}\begin{align} \sigma_2 = -5.65 \times 10^{11} \ \rm N/C \times 2 \times 8.85 \times 10^{-12} \ \rm F/m \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ \ \sigma_2 = -10.0 \ \rm C/m^2 \ }} \end{align} {/eq}


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