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

Answer:

$Ep_x = 288.97*10^3\frac{N}{C}$

$Ep_y = 2770.6*10^3\frac{N}{C}$

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

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

$\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o$

Problem development

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

$Ep = Ep_x i + Ep_y j$

Ep₁ₓ = 0

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

$Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}$

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

Calculation of the electric field components at point P

$Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}$

$Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}$