Question

Q11. Compute electric potential and potential gradient at a distance of 1.0*10^-12 m from the centre of a gold nucleus. Hence find e-field at that point. The atomic number of gold is 79​

Answer 1

Given:

Gold nucleus with atomic number 79.

To find:

• Electric potential
• Potential Gradient
• Electric field intensity

Calculation:

So, net charge in nucleus is :

$q = ne$

$\implies q = 79 \times (1.6 \times {10}^{ - 19} )$

$\implies q = 126.4 \times {10}^{ - 19} \: C$

Now, electric potential :

$V = \dfrac{kq}{r}$

$\implies V = \dfrac{9 \times {10}^{9} \times 126.4 \times {10}^{ - 19} }{ {10}^{ - 12} }$

$\implies V = 1137.6\times {10}^{2}$

$\implies V = 1.14\times {10}^{5} \: volt$

Potential Gradient is :

$gradient = \dfrac{V}{r}$

$\implies gradient = \dfrac{1.14 \times {10}^{5} }{ {10}^{ - 12} }$

$\implies gradient = 1.14 \times {10}^{17} \: volt \: {m}^{ - 1}$

Electric Field Intensity:

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

$\implies E = \dfrac{9 \times {10}^{9} \times 126.4 \times {10}^{ - 19} }{{ ({10}^{ - 12})}^{2} }$

$\implies E = \dfrac{9 \times {10}^{9} \times 126.4 \times {10}^{ - 19} }{ {10}^{ - 24} }$

$\implies E = 1137.6 \times {10}^{14} \: N/C$

Hope It Helps.