Question

A charge of 63.0 nC is located at a distance of 3.40 cm from a charge of -47.0 nC. What are the x- and y-components of the electric field at a point P that is directly above the 63.0 nC charge at a distance of 1.40 cm? Point P and the two charges are on the vertices of a right triangle.

Answer 1

`Ep_x = 288.97*10^3(N)/(C)`

`Ep_y = 2770.6*10^3(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

1nC= 10⁻⁹ 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 positive ,the field leaves the charge.

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

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

`r=√(d_1^2 + d_2^2) = \sqrt{(1.4*10^(-2))^2 + (3.4*10^(-2))^2} = 3.677*10^(-2)m`

`\beta = tan^(-1)((d_1)/(d_2)) = tan^(-1)((1.4)/(3.4)) = 22.38^o`

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

`Ep = Ep_x i + Ep_y j`

Ep₁ₓ = 0

`Ep_(2x)=(-k*q_2*Cos\beta)/(r^2)=(8.99*10^9*47*10^(-9)*Cos(22.38))/((3.677*10^(-2))^2)=288.97*10^3(N)/(C)`

`Ep_(1y)=(-k*q_1)/(d_1^2)=(8.99*10^9*63*10^(-9))/((1.4*10^(-2))^2)=2889.6*10^3(N)/(C)`

`Ep_(2y)=(-k*q_2*Sen\beta)/(r^2)=(-8.99*10^9*47*10^(-9)*Sen(22.38))/((3.677*10^(-2))^2)=-119*10^3(N)/(C)`

Calculation of the electric field components at point P

`Ep_x = Ep_(1x) + Ep_(2x) = 0 + 288.97*10^3 = 288.97*10^3(N)/(C)`

`Ep_y = Ep_(1y) + Ep_(2y) = 2889.6*10^3 - 119*10^3 = 2770.6*10^3(N)/(C)`