Question

An atomic nucleus has a charge of +40e. What is the magnitude of the electric field at a distance of 1.0 m from the center of the nucleus?

Answer 1

Answer:

$E=5.76\times 10^{-8}N/C$

Step-by-step explanation:

We are given that

Charge of atomic nucleus,q=+40 e=$40\times 1.6\times 10^{-19}C$

Because

1 e=$1.6\times 10^{-19} C$

Charge of atomic nucleus=$64\times 10^{-19}C$

Distance of charge form the center of nucleus=r=1 m

We have to find the magnitude of  electric field at distance r

$E=\frac{Kq}{r^2}$

Where K=$9\times 10^9Nm^2/C^2$

Using the formula

$E=\frac{9\times 10^9\times 64\times 10^{-19}}{1}=576\times 10^{-10}N/C$

$E=\frac{576}{100}\times 10^2\times 10^{-10}=5.76\times 10^{2-10}=5.76\times 10^{-8}N/C$

Using identity$a^x\cdot a^y=a^{x+y}$

Hence, the magnitude of electric field=$E=5.76\times 10^{-8}N/C$