Physics, 26.09.2019 16:30 mjohn6150

Two charges, one +e and one −e, are placed on an x-axis, separated by a distance d = 12µm. exactly halfway between the two charges, and a height h = 5µm above the x-axis is a point p. what is the magnitude and direction of the net electric field at p?

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Answer from: jasminecoronetti44

Ep= 226.68*e*10¹⁸ N/C in direction (+x)

Explanation:

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1µm= 10⁻⁶m

Data

k= 9*10⁹ N*m²/C²

q₁= e

q₂= -e

d₁=d₂=d

$d=\sqrt{6^{2}+5^{2} } *10^{-6} m$

d=7,81µm=7,81*10⁻⁶m

Calculation of the electric fields at the point due to q₁ and q₂

Look at the attached graphic:

E₁: Electric Field at point P due to charge q₁. As the charge q₁ is positive (e+) ,the field leaves the charge

E₂: Electric Field at point due to charge q₂. As the charge q₂ is negative (e-), the field enters the charge

Since the magnitude of q₁ is equal to the magnitude of q₂ and the distance of q₁ to the point P is equal to the distance of q₂ to the point P, then, the magnitude of E₁ is equal to the magnitude of E₂:

$E=E_{1} =E_{2} =\frac{k*q}{d^{2} }$

E=9*10⁹ *e/ (7.81 *10⁻⁶)² =147.5*e*10¹⁸ N/C

E₁x=E₂x=Ecosβ=147.5*e*10¹⁸ *(6/7.81)= +113.34*e*10¹⁸ N/C

E₁y=Esinβ=147.5*e*10¹⁸ *(5/7.81)= +94.43*e*10¹⁸ N/C

E₂y= -Esinβ=-147.5*e*10¹⁸ *(5/7.81)= -94.43*e*10¹⁸ N/C

Components x-y of the electric field at point P due to q₁ and q₂

Epx = E₁x+E₂x =+113.34*e*10¹⁸ +113.34*e*10¹⁸ =+ 226.68*e*10¹⁸ N/C

Epy = E₁y+E₂y =+94.43*e*10¹⁸ - 94.43*e*10¹⁸ = 0

Magnitude and direction of the net electric field at P

Ep=Epx=+ 226.68*e*10¹⁸ N/C in direction (+x)

Two charges, one +e and one −e, are placed on an x-axis, separated by a distance d = 12µm. exactly h

Answer from: Quest

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Answer from: Quest

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