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

inside a television picture tube there is a build - up of electrons (charge of 1.602×10⁻¹⁹ c) with an average spacing of 38.0 nanometers between electrons. find the electric field inside the picture tube. (round to three significant figures.)

inside a television picture tube there is a build - up of electrons (charge of 1.602×10⁻¹⁹ c) with an average spacing of 38.0 nanometers between electrons. find the electric field inside the picture tube. (round to three significant figures.)

Answer

Explanation:

Step1: Recall electric - field formula

The electric field due to a point - charge is given by $E = \frac{kq}{r^{2}}$, where $k = 8.99\times10^{9}\ N\cdot m^{2}/C^{2}$, $q$ is the charge, and $r$ is the distance from the charge. Here, we can consider the electric field due to one electron at the position of another electron. The charge of an electron $q = 1.602\times10^{-19}\ C$ and the distance between electrons $r=38.0\ nm = 38.0\times10^{-9}\ m$.

Step2: Substitute values into the formula

Substitute $k = 8.99\times10^{9}\ N\cdot m^{2}/C^{2}$, $q = 1.602\times10^{-19}\ C$, and $r = 38.0\times10^{-9}\ m$ into the formula $E=\frac{kq}{r^{2}}$. [ \begin{align*} E&=\frac{8.99\times 10^{9}\times1.602\times 10^{-19}}{(38.0\times 10^{-9})^{2}}\ &=\frac{8.99\times1.602\times 10^{9 - 19}}{38.0^{2}\times10^{-18}}\ &=\frac{14.40198\times10^{-10}}{1444\times10^{-18}}\ &=\frac{14.40198}{1444}\times10^{-10 + 18}\ &=0.00997367\times10^{8}\ & = 9.97\times10^{5}\ N/C \end{align*} ]

Answer:

$9.97\times10^{5}$