Two point charges (q1 = -2.5μc and q2 = 7.2 μc) are fixed along the x-axis, separated by a distance d = 9.9 cm. point p is located at (x, y) = (d, d). be sure to use a vector approach to answer these problems. it also limit the number of calculations since sometimes you can neglect one of the charge's effects due to its location relative to the point in question. 1) what is ex(p), the value of the x-component of the electric field produced by q1 and q2 at point p? n/c 2) what is ey(p), the value of the y-component of the electric field produced by q1 and q2 at point p?

Epx = -0.81*10⁶ N/C

Epy = 5.79*10⁶ N/C

Explanation:

Electric field theory

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 load 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.

Known data

q₁ = -2.5 µC = -2.5*10⁻⁶ C

q₂ = 7,2 µC = 7,2*10⁻⁶ C

k = 8.99*10⁹ N*m²/C²

d = 9.9cm = 9.9*10⁻² m

θ = 45°

sinθ = cosθ =

r calculation

Calculation of the electric field at point P due to q₁

Ep1x = (-k*q₁*cosθ)/r²

Ep1y = (-k*q₁*sinθ)/r²

Calculation of the electric field at point P due to q₂

Ep2x=0

Ep2y=k*q₁/r²=(8.99*10⁹*7.2 *10⁻⁶)/

(9.9*10⁻²)² = 6.6*10⁶ N/C

Calculation of the electric field at point P due to q₁ and q₂

Epx = Ep1x + Ep2x = -0.81*10⁶ + 0 = -0.81*10⁶ N/C

Epy = Ep1y+ Ep2y= -0.81*10⁶ + 6.6*10⁶ = 5.79*10⁶ N/C