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Charge of uniform linear density (4.0 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 2.5 m

Respuesta :

okpalawalter8

Answer:

The Magnitude of electric field is  [tex]E =29 N/C[/tex]

Explanation:

From the question we are told that

     The  uniform linear density is [tex]\lambda = 4.0nC/m[/tex]

      The distance on the y-axis is [tex]d = 2.5m[/tex]

Generally electric field is mathematically represented as

                                [tex]E =\frac{\lambda}{2 \pi r \epsilon_o d }[/tex]

Where [tex]\epsilon_o[/tex] is the permitivity of free space given as [tex]\epsilon_0 = 8.85*10^{-12} C^2 /N \cdot m^2[/tex]

Substituting values

                          [tex]E= \frac{1}{2} * 3.142 * 8.85*10^{-12} * \frac{4.0*10^{-9}}{2.5}[/tex]

                               [tex]= 29N/C[/tex]

Cricetus

On the y-axis at y = 2.5 m, the magnitude of the electric field will be "29 N/C".

Electric field

According to the question,

Uniform linear density, λ = 4.0 nC/m

y-axis distance, d = 2.5 m

We know the relation,

Electric field, E = [tex]\frac{\lambda}{2 \pi r \epsilon_0 d}[/tex]

By substituting the values in the above relation,

                         = [tex]\frac{1}{2}[/tex] × 3.142 × 8.85 × 10⁻¹² × [tex]\frac{4.0\times 10^{-9}}{2.5}[/tex]

                         = [tex]\frac{1}{2}[/tex] × 27.8067 × 10⁻¹² × [tex]\frac{4.0\times 10^{-9}}{2.5}[/tex]

                         = 29 N/C

Thus the above approach is correct.                        

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