Respuesta :
Explanation:
The electric field at a distance r from the charged particle is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
k is electrostatic constant
if r = 2 m, electric field is given by :
[tex]E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)[/tex]
If r = 1 m, electric field is given by :
[tex]E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)[/tex]
Dividing equation (1) and (2) we get :
[tex]\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1[/tex]
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
At a point 1 m from the particle, the magnitude of the electric field is;
4 times the magnitude of the electric field is at a distance of 2 m from the particle.
Formula for magnitude of electric field is;
E = kq/r²
where;
k is a constant = 8.99 x 10⁹ N.m²/C²
r is distance from the charged particle
q is the charge
Now, when r = 2, we have;
E₁ = kq/2²
E₁ = kq/4 ---(eq 1)
Now, when r = 1, we have;
E₂ = kq/1²
E₂ = kq ---(eq 2)
Let us now divide eq 2 by eq 1 to get;
E₂/E₁ = kq/(kq/4)
E₂/E₁ = 4
E₂ = 4E₁
Thus, at a point 1 m from the particle, the magnitude of the electric field is 4 times the magnitude of the electric field is at a distance of 2 m from the particle.
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