inside a television picture tube there is a build - up of electrons (charge of 1.602×10^(-19) c) with an…

Explanation:

Step1: Recall the formula for electric - field due to a point - charge

The electric field due to a point - charge $q$ at a distance $r$ is given by $E = \frac{kq}{r^{2}}$, where $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$, $q$ is the charge, and $r$ is the distance from the charge.

Step2: Convert the distance to SI units

The average spacing between electrons $r = 38.0\ nm=38.0\times10^{-9}\ m$, and the charge of an electron $q = 1.602\times10^{-19}\ C$.

Step3: Substitute the values into the formula

Substitute $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$, $q = 1.602\times10^{-19}\ C$, and $r = 38.0\times10^{-9}\ m$ into the formula $E=\frac{kq}{r^{2}}$. [ \begin{align*} E&=\frac{(9.0\times 10^{9}\ N\cdot m^{2}/C^{2})\times(1.602\times 10^{-19}\ C)}{(38.0\times 10^{-9}\ m)^{2}}\ &=\frac{9.0\times1.602\times 10^{9 - 19}\ N\cdot m^{2}/C}{38.0^{2}\times10^{-18}\ m^{2}}\ &=\frac{14.418\times10^{-10}\ N\cdot m^{2}/C}{1444\times10^{-18}\ m^{2}}\ &=\frac{14.418}{1444}\times10^{-10 + 18}\ N/C\ &=0.01\times10^{8}\ N/C\ & = 1.00\times10^{6}\ N/C \end{align*} ]

Answer:

$1.00\times 10^{6}$