Answer:
The magnitude of the electric field in the air gap [tex]E = 0.00036 C[/tex]
Explanation:
The Electric field E between the plates, [tex]E = \frac{q}{4\pi \epsilon_{0} r^{2} }[/tex]
Where q = the positive charge
r = separation of the plates= 0.002 m
[tex]\frac{1}{4\pi \epsilon_{0} } = 9 * 10^{9} Nm^{2} /C^{2}[/tex]
[tex]E = \frac{9 * 10^{9} q}{0.002^{2} } \\E = \frac{9 * 10^{9} q}{4 * 10^{-6} } \\E = 2.25* 10^{15} q[/tex]
The elementary positive charge, q = 1.602176634×10−19 C
[tex]E = 2.25 * 10^{15} * 1.602176634×10^{-19} \\E = 0.00036 C[/tex]
