Answered

an interstellar dust grain, roughly spherical with a radius of 3 ×10−7 m, has acquired a negative charge such that its potential is −0.15 volt. how many extra electrons has it picked up? what is the strength of the electric field at its surface?



Answer :

  1. Number of extra electrons picked up by the dust grain = 31
  2. The strength of the electric field at its surface = 5 * [tex]10^{5}[/tex] V / m

V = k q / d

V = Electric potential difference

k = Coulomb's constant

q = Electric charge

d = Distance from the point charge

Since it is spherical in shape, according to Gauss law, the dust grain can be treated as a point charge.

[tex]q_{net}[/tex] = - n e

[tex]q_{net}[/tex] = Net charge

n = No. of excess electrons

e = Charge of an electron

V = - 0.15 V

d = 3 * [tex]10^{-7}[/tex] m

k = 9 * [tex]10^{9}[/tex] N

e = 1.6 * [tex]10^{-19}[/tex] C

[tex]q_{net}[/tex] = - n ( 1.6 * [tex]10^{-19}[/tex] )

V = k [tex]q_{net}[/tex] / d

- 0.15 = - n ( 9 * [tex]10^{9}[/tex] ) ( 1.6 * [tex]10^{-19}[/tex] ) / 3 * [tex]10^{-7}[/tex]

n = 31 electrons

E = k q / r²

E = Electric field strength

E = - k n e / r²

E = - ( 9 * [tex]10^{9}[/tex] ) ( 31 ) ( 1.6 * [tex]10^{-19}[/tex] ) / ( 3 * [tex]10^{-7}[/tex] )²

E = 5 * [tex]10^{5}[/tex] V / m

Therefore,

  1. Number of extra electrons picked up by the dust grain = 31
  2. The strength of the electric field at its surface = 5 * [tex]10^{5}[/tex] V / m

To know more about number of excess electrons

https://brainly.com/question/17354475

#SPJ1

Other Questions