Question

Dry air will break down if the electric field exceeds about3.0 × 10⁶ V/m. What amount of charge can be placedon a capacitor if the area of each plate is 3.1 cm²?

                  nC

I tried using this formula rearranged for Q :

Q = Eε₀A

    = (3.0 × 10⁶ V/m)(8.85 x10^-12 Nm²/C²)(.031m²) = 8.2305 x 10^-7C = 8.2405e-7nC

However, this is not right. Can anyone tell me what I amdoing wrong?

Answer 1

Concepts and reason

The concepts used to solve this problem are charge stored on the plates of capacitor and

Electric field for a parallel plate capacitor.

Initially, use the equation of electric field between the plates of capacitor, re-arrange the equation for Q, substitute the values and find the result.

Fundamentals

The electric field between parallel plates of a parallel plate capacitor is given as follows:

$E = \frac{\sigma }{{{\varepsilon _0}}}$

Here, $\sigma$ is the surface charge density and ${\varepsilon _0}$ is the electrical permittivity of vacuum.

The surface charge density is given as follows:

$\sigma = \frac{Q}{A}$

Here, Q is the charge stored and A is the area.

Substitute equation $\sigma = \frac{Q}{A}$ in equation $E = \frac{\sigma }{{{\varepsilon _0}}}$ and re-arrange the expression for Q.

$\begin{array}{c}\\E = \frac{\sigma }{{{\varepsilon _0}}}\\\\ = \frac{Q}{{A{\varepsilon _0}}}\\\\Q = AE{\varepsilon _0}\\\end{array}$

Substitute $3.0 \times {10^6}{\rm{ V/m}}$ for E, $3.1{\rm{ c}}{{\rm{m}}^2}$ for A and $8.85 \times {10^{ - 12}}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}$ for ${\varepsilon _0}$ in equation $Q = AE{\varepsilon _0}$ .

$\begin{array}{c}\\Q = \left( {3.1{\rm{ c}}{{\rm{m}}^2}\left( {\frac{{{{10}^{ - 4}}{\rm{ }}{{\rm{m}}^2}}}{{1.0{\rm{ c}}{{\rm{m}}^2}}}} \right)} \right)\left( {3.0 \times {{10}^6}{\rm{ V/m}}} \right)\left( {8.85 \times {{10}^{ - 12}}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\\\\ = 8.23 \times {10^{ - 9}}{\rm{ C}}\\\\ = 8.23{\rm{ nC}}\\\end{array}$

Ans:

The amount of charge stored is 8.23 nC.