Answer:
The electric force on an electron in this field [tex]3.2\times 10^{-15}\ N[/tex].
Explanation:
We have,
Electric field of along the +x-axis is [tex]2\times 10^4\ N/C[/tex]
It is required to find the electric force on an electron in this field. The electric force acting on the particle is given by :
[tex]F=qE[/tex]
q is charge of electron
[tex]q=-1.6\times 10^{-19}\ C[/tex]
So,
[tex]F=1.6\times 10^{-19}\times 2\times 10^4\\\\F=3.2\times 10^{-15}[/tex]
So, the electric force on an electron in this field [tex]3.2\times 10^{-15}\ N[/tex].
The electric force on an electron throughout this field will be "3.2 × 10⁻¹⁵ N".
An electric field throughout a force created by a charge throughout proximity to its circumstances. Whenever additional charges come into contact with this field, they are subjected to this force.
A charge's electric field at quite a point seems to be the force that somehow a unit positive (+) charge will indeed personal experience if put at around that spot.
According to the question,
Electric field = 2 × 10⁴ N/C
Charge of electron, q = -1.6 × 10⁻¹⁹ C
We know the relation,
→ F = qE
By substituting the values, we get
= [tex]1.6\times 10^{-19}\times 2\times 10^4[/tex]
= [tex]3.2\times 10^{-15}[/tex] N
Thus the above answer is correct.
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