Question

Two point charges (q1 = -3.6μC and q2 = 8.5 μC) are fixed along the x-axis, separated by a distance d = 6.8 cm. Point P is located at (x,y) = (d,d).1) What is Ex(P), the value of the x-component of the electric field produced by q1 and q2 at point P?

Answer 1

Epₓ=178.78×10⁶ N/C

Step-by-step explanation:

Conceptual analysis

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

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

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

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1µC= 10⁻⁶ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is negative, the field enters the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is positive ,the field leaves the charge.

Ep₁ₓ: Total field at point P due to charge q₁ in the x direction.

Ep₂ₓ: Total field at point P due to charge q₂ in the x direction.

Known data

q₁ = -3.6 µC = -3.6×10⁻⁶ C

q₂ = 8.5 µC = 8.5×10⁻⁶ C

k = 8.99*10⁹ N*m²/C²

d = 6.8cm = 6.8*10⁻² m

Calculation of r and β

`r=√(0.8^2+0.8^2) * 10^(-2)m= √(1.28) * 10^(-2)m`

`\beta = tan^(-1) ((0.8)/(0.8))= tan^(-1) (1)=45^o`

Problem development

Epₓ: Total field at point P due to charges q₁ and q₂ in the x direction.

Epₓ = Ep₁ₓ + Ep₂ₓ

Ep₂ₓ = 0

Ep₁ₓ = (-k×q₁×Cosβ)/r²

Epₓ = Ep₁ₓ + 0

Epₓ = Ep₁ₓ

`Ep_x=(8.99*10^9*3.6*10^(-6)*Cos(45))/((√(1.28)*10^(-2))^2 ) = 178.78*10^6(N)/(C)`