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Question

Air “breaks down” when the electric field strength reaches 3.0×106N/C3.0 \times 10^{6} \mathrm{N} / \mathrm{C}, causing a spark. A parallel-plate capacitor is made from two 4.0cm×4.0cm4.0 \mathrm{cm} \times 4.0 \mathrm{cm} electrodes. How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?

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The electric field strength in a parallel capacitor is given by the formula

E=σε0,E=\frac{\sigma}{\varepsilon_0},

where one electrode carries the +σ+\sigma surface charge density and the other one carries σ-\sigma surface charge. In the beginning the electrodes are neutral so the field is zero. We can create σ-\sigma on one electrode and +σ+\sigma in the other one by tranfering electrons from one another. Each electron carries the charge of magnitude ee. So by trasfering NN electrons we transfer the total charge of NeNe and this creates the surface charge density of

σ=Nea2,\sigma=\frac{Ne}{a^2,}

where a=4.0a=4.0 cm. This yields

E=Nea2ε0.E=\frac{Ne}{a^2\varepsilon_0}.

Solving for NN we find

N=a2ε0Ee.N=\frac{a^2\varepsilon_0 E}{e}.

Putting in E=3.0106E=3.0\cdot10^6 N/C we get

N=2.71011.\boxed{N=2.7\cdot10^{11}.}

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