Question

In Problems 69 through 72 you are given the equation(s) used to solve a problem. For each of these,

1. Write a realistic problem for which this is the correct equation(s).
2. Finish the solution of the problem.

$$\begin{gathered} \frac{(9.0\times {10}^{9}N{m}^2/C^2)\times N\times (1.60\times {10}^{-19}C)}{{(1.0\times {10}^{-6}m)}^2} \\ =1.5\times {10}^{6}N/C \end{gathered}$$

Short answer

1. The electric field due to a point charge at distance of $1.0\mu \mathrm{m}$is given as $1.5\times {10}^{6}N/C$. Calculate the number of electrons within the point charge.
2. The number of electrons is$1.0\times {10}^3$.

Step-by-step solution

Part (a) Step 1: Given Information

Equation $=\frac{(9.0\times {10}^{9}N{m}^2/C^2)\times N\times (1.60\times {10}^{-19}C)}{{(1.0\times {10}^{-6}m)}^2}=1.5\times {10}^{6}N/C$

Part (a) Step 2: Explanation

By comparing the given equation from the equation of the magnitude of the electric field by a point charge which is given as:

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

We get,

Electric field $=1.5\times {10}^{6}N/C$

Distance from the point charge $=1.0\times {10}^{-6}m=1.0\mu m$

Constant localid="1648625943877" $=\frac{1}{4\pi {ϵ}_0}=9.0\times {10}^{9}N{m}^2/C^2$

Amount of charge $=N\times 1.60\times {10}^{-19}C$

Where,

$N=$number of electrons

$1.60\times {10}^{-19}C=$charge of one electron

Hence, it can be concluded that an electric field of$1.5\times {10}^{6}N/C$is generated by$N$electrons at a distance of$1.0\mu m$.

Part (a) Step 3: Final answer

Hence, the realistic problem that can be generated for the given question is:

The electric field due to a point charge at distance of $1.0\mu m$is given as $1.5\times {10}^{6}N/C$. Calculate the number of electrons within the point charge.

Part (b) Step 1: Given Information

Equation$=\frac{(9.0\times {10}^{9}N{m}^2/C^2)\times N\times (1.60\times {10}^{-19}C)}{{(1.0\times {10}^{-6}m)}^2}=1.5\times {10}^{6}N/C$

Part (b) Step 2: Calculation

The given equation is:

$\frac{(9.0\times {10}^{9}N{m}^2/C^2)\times N\times (1.60\times {10}^{-19}C)}{{(1.0\times {10}^{-6}m)}^2}=1.5\times {10}^{6}N/C$

Based on the equation, a realistic problem can be made as:

The electric field due to a point charge at distance of $1.0\mu m$is given as$1.5\times {10}^{6}N/C$. Calculate the number of electrons within the point charge.

Hence, by rearranging the terms to get the value of N,

$$\begin{gathered} N=\frac{1.5\times {10}^{6}N/C\times {(1.0\times {10}^{-6}m)}^2}{(9.0\times {10}^{9}N{m}^2/C^2)\times (1.60\times {10}^{-19}C)} \\ N=1041.6\approx 1.0\times {10}^3 \end{gathered}$$

Part (b) Step 3: Final answer

Hence, the number of electrons is calculated to be$1.0\times {10}^3$.