Question

A battery connected across two parallel metal plates. There is a uniform E-field between the plates, and a positive charge experiences a drop in potential upon traveling from the left plate to the right plate. If the separation of the plates is 0.002 m, determine the magnitude of the electric field in the air gap

Answer 1

Answer:

The magnitude of the electric field in the air gap $E = 0.00036 C$

Explanation:

The Electric field E between the plates, $E = \frac{q}{4\pi \epsilon_{0} r^{2} }$

Where q = the positive charge

r = separation of the plates= 0.002 m

$\frac{1}{4\pi \epsilon_{0} } = 9 * 10^{9} Nm^{2} /C^{2}$

$E = \frac{9 * 10^{9} q}{0.002^{2} } \\E = \frac{9 * 10^{9} q}{4 * 10^{-6} } \\E = 2.25* 10^{15} q$

The elementary positive charge, q = 1.602176634×10−19 C

$E = 2.25 * 10^{15} * 1.602176634×10^{-19} \\E = 0.00036 C$