Question

What is the diameter of sphere of 4V potential kept near the points situated at same distance

from an electron ?​

Answer 1

Correct Question and Options have been attached !

Calculation:

Now, the potential at the surface of the sphere is equal to the potential of the electron at that distance.

And the radius of sphere is equal to the distance of that point from the sphere

$\therefore \: V = \dfrac{q_{s}}{4\pi\epsilon_{0} r}$

$\implies \: V = \dfrac{q_{e}}{4\pi\epsilon_{0} r}$

$\implies \: r = \dfrac{q_{e}}{4\pi\epsilon_{0}V}$

$\implies \: d =2r = 2 \times \dfrac{q_{e}}{4\pi\epsilon_{0}V}$

$\implies \: d = 2 \times \dfrac{q_{e}}{4\pi\epsilon_{0}V}$

$\implies \: d = 2 \times \dfrac{(1.6 \times {10}^{ - 19} ) \times (9 \times {10}^{9}) }{4}$

$\implies \: d = \dfrac{(1.6 \times {10}^{ - 19} ) \times (9 \times {10}^{9}) }{2}$

$\implies \: d = 7.2\times {10}^{ - 10}$

$\implies \: d = 7.2 A^{\circ}$

So, option B) is Correct .

Answer 2

${\bold{\green{\underline{\underline{Given...}}}}}$

Potential ⟼ 4V

$Potential = \frac{ke}{r}$

here e = Electron

$4 = \frac{ke}{r} \\ \\ r = \frac{ke}{4}$

$Diameter = 2 \times radius$

$⇒ \frac{ke}{4}$

$\;\large{\boxed{\bf{\red{value \: of \: (k) = 9 \times 10 {}^{9} }}}} \\ \;\large{\boxed{\bf{\red{value \: of \: (e) = 1.6 \times 10 {}^{ - 19} }}}}$

After Calculation ,

$\;\large{\boxed{\bf{\purple{7.2 \times 10 {}^{ - 10 \: } }}}} \: is \: th \: ans...$

$\sf\blue{hope \: this \: helps \: you!! \: }$

@Mahor2111