Question

Two charges are located on the x-axis. A 22 µC charge is located at the origin and a 47 µC charge is located at x = 4.0 m. What is the magnitude of the electric field at x = 7.5 m?

Answer 1

Explanation:

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Answer 2

Answer:

Two charges are located on the x-axis. A 22 µC charge is located at the origin and a 47 µC charge is located at x = 4.0 m, the magnitude of the electric field at x = 7.5 m is $38*10^{3}N/C$

Explanation:

• The electric field due to a point charge (q) at a distance (r) from the charge is given by the formula

       $E=K\frac{q}{r^{2} }$

      where the constant $K=9*10^{9}$

From the question, we have an arrangement as in the figure

The electric field due to the first charge at point P is

$E_{1}=\frac{9*10^{9}*22*10^{-6} }{7.5^{2} }$

⇒ $E_{1} =3.5*10^{3}N/C$

The electric field due to the second charge at point P is

$E_{2}=\frac{9*10^{9}*47*10^{-6} }{3.5^{2} }$

$E_{2} =34.6*10^{3}N/C$

since both, the fields are directed along the same direction net electric field at P is

$E_{net} =E_{1} +E_{2} =(3.5+34.6)*10^{3}=38*10^{3} N/C$