Question

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

Answer 1

Explanation:

We are given that

Charge of atomic nucleus,q=+40 e=40\times 1.6\times 10^{-19}C40×1.6×10−19C

Because

1 e=1.6\times 10^{-19} C1.6×10−19C

Charge of atomic nucleus=64\times 10^{-19}C64×10−19C

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}E=r2Kq

Where K=9\times 10^9Nm^2/C^29×109Nm2/C2

Using the formula

E=\frac{9\times 10^9\times 64\times 10^{-19}}{1}=576\times 10^{-10}N/CE=19×109×64×10−19=576×10−10N/C

E=\frac{576}{100}\times 10^2\times 10^{-10}=5.76\times 10^{2-10}=5.76\times 10^{-8}N/CE=100576×102×10−10=5.76×102−10=5.76×10−8N/C

Using identitya^x\cdot a^y=a^{x+y}ax⋅ay=ax+y

Hence, the magnitude of electric field=

Answer 2

The magnitude of the electric field at a distance of 1 m from the center of the nucleus is $5.76\times 10^{-8}\ N/C$.

Explanation:

Charge on an atomic nucleus, $q=40e=40\times 1.6\times 10^{-19}=6.4\times 10^{-18}\ C$

It is required to find the magnitude of the electric field at a distance of 1 m from the center of the nucleus. The electric field is given by :

$E=\dfrac{kq}{r^2}$

k is electric constant

$E=\dfrac{9\times 10^9\times 6.4\times 10^{-18}}{1^2}\\\\E=5.76\times 10^{-8}\ N/C$

So, the magnitude of the electric field at a distance of 1 m from the center of the nucleus is $5.76\times 10^{-8}\ N/C$.

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