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Calculate the electric field at the center of a square
52.5cmon a side if one corner is occupied by a +45.0uC charge and
theother three are occupied by -27.0uC charges?

Respuesta :

Answer:

[tex]E_y=1175510.2\ N.C^{-1}[/tex]

The Magnitude of electric field is in the upward direction as shown directly towards the charge [tex]q_1[/tex].

Explanation:

Given:

  • side of a square, [tex]a=52.5\ cm[/tex]
  • charge on one corner of the square, [tex]q_1=+45\times 10^{-6}\ C[/tex]
  • charge on the remaining 3 corners of the square,[tex]q_2=q_3=q_4=-27\times 10^{-6}\ C[/tex]

Distance of the center from each corners[tex]=\frac{1}{2} \times diagonals[/tex]

[tex]diagonal=\sqrt{52.5^2+52.5^2}[/tex]

[tex]diagonal=74.25\cm=0.7425\ m[/tex]

∴Distance of center from corners, [tex]b=0.3712\ m[/tex]

Now, electric field due to charges is given as:

[tex]E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}[/tex]

For charge [tex]q_1[/tex] we have the field lines emerging out of the charge since it is positively charged:

[tex]E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}[/tex]

  • [tex]E_1=2938775.5\ N.C^{-1}[/tex]

Force by each of the charges at the remaining corners:

[tex]E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}[/tex]

  • [tex]E_2=E_3=E_4=1763265.3\ N.C^{-1}[/tex]

Now, net electric field in the vertical direction:

[tex]E_y=E_1-E_4[/tex]

[tex]E_y=1175510.2\ N.C^{-1}[/tex]

Now, net electric field in the horizontal direction:

[tex]E_y=E_2-E_3[/tex]

[tex]E_y=0\ N.C^{-1}[/tex]

So the Magnitude of electric field is in the upward direction as shown directly towards the charge [tex]q_1[/tex].

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