Question

A charge of \(63.0 \mathrm{nC}\) is located at a distance of \(3.40 \mathrm{cm}\) from a charge of \(-47.0 \mathrm{nC} .\) 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 \(1.40 \mathrm{cm} ?\) Point \(P\) and the two charges are on the vertices of a right triangle.

Short answer

Answer: The x-component of the electric field at point P is \(E_x = -3.27 x 10^5\) N/C, and the y-component is \(E_y = 0.54 x 10^5\) N/C.

Step-by-step solution

Convert units and write down known values

: To calculate the electric field, we need to use the Coulomb’s law formula. To do that, let's first convert the given values into SI units. Charge 1, \(q_1 = 63.0\) nC \(= 63.0 x 10^{-9}\) C Charge 2, \(q_2 = -47.0\) nC \(= -47.0 x 10^{-9}\) C Distance between the charges, \(a = 3.40\) cm \(= 0.034\) m Height above the 63.0 nC charge (vertical distance), \(b = 1.40\) cm \(= 0.014\) m

Calculate the electric field due to the 63.0 nC charge

: Since point P is directly above the 63.0 nC charge, the electric field due to this charge is in the vertical direction (y-direction): E_y due to q1: \(E_{y1} = \frac{kq_1}{b^2}\) Where \(k = 8.99 x 10^9\) Nm²/C² is the electrostatic constant. Calculating \(E_{y1}\): \(E_{y1} = \frac{(8.99 x 10^9)(63.0 x 10^{-9})}{(0.014)^2} = 2.01 x 10^5\) N/C

Calculate the electric field due to the -47.0 nC charge

: Since point P is horizontally 3.4 cm away from the -47.0 nC charge, we can use geometry to find the angle between the horizontal (x-direction) and electric field vector and then find both components of the electric field due to this charge. First, find the total electric field due to q2: \(E_2 = \frac{kq_2}{a^2 + b^2}\) Calculating \(E_2\): \(E_2 = \frac{(8.99 x 10^9)(-47.0 x 10^{-9})}{(0.014^2 + 0.034^2)} = -3.61 x 10^5\) N/C Now, find the angle θ between the horizontal (the x-axis) and the electric field vector: \(sin(\theta) = \frac{b}{\sqrt{a^2 + b^2}}\) Calculating θ using the inverse sine function and converting to degrees (optional): \(\theta = arcsin(\frac{0.014}{\sqrt{0.014^2 + 0.034^2}}) = 22.81°\) Finally, find electric field components due to q2: \(E_{x2} = E_2cos(\theta) \) \(E_{y2} = E_2sin(\theta) \) Calculating \(E_{x2}\) and \(E_{y2}\): \(E_{x2} = (-3.61 x 10^5)cos(22.81°) = -3.27 x 10^5\) N/C \(E_{y2} = (-3.61 x 10^5)sin(22.81°) = -1.47 x 10^5\) N/C

Find total electric field components at P

: Finally, we add up the electric field components due to both charges: \(E_x = E_{x2} = -3.27 x 10^5\) N/C \(E_y = E_{y1} + E_{y2} = 2.01 x 10^5 + (-1.47 x 10^5) = 0.54 x 10^5\) N/C

Final answer

: The x- and y-components of the electric field at point P are: \(E_x = -3.27 x 10^5\) N/C \(E_y = 0.54 x 10^5\) N/C