## Related questions with answers

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

Solution

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

$E=\frac{\sigma}{\varepsilon_0},$

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

$\sigma=\frac{Ne}{a^2,}$

where $a=4.0$ cm. This yields

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

Solving for $N$ we find

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

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

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

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Introduction to Electrodynamics

4th Edition•ISBN: 9780321856562 (3 more)David J. Griffiths#### Introduction to Quantum Mechanics

3rd Edition•ISBN: 9781107189638Darrell F. Schroeter, David J. Griffiths## More related questions

1/4

1/7