a positive charge of magnitude Q1 = 0.95nC is located at the origin

Question

a positive charge of magnitude Q1 = 0.95nC is located at the origin. A negative charge Q2 = -8.5nC is located on the positive x-axis at x=11cm from the origin. The point P is located y =17cm above charge Q2.

part a) Calculate the y-component of the electric field at point P due to the charge Q2.

part b) Calculate the y-component of the electric field at point P due to BOTH charges.

part c) Calculate the magnitude of the electric field at point P due to BOTH charges.

part d) Calculate the angle in degrees of the electric field at point P relative to the positive x-axis.

Answer 1

Hello CX S,

I see that you posted this question 10 days ago. It was just sent to me today. So hopefully this still helps.

a) The Electric Field at point P due to Charge Q2 is given by the equation

E_Q2 = k q/r² = (8.99 x 10⁹ N m² /C²) (-8.5 x 10^-9 C)/(17x10^-2 m)² = -2644 N/C (the - indicates the field is in the -y direction. Since Q2 is directly below point P on the coordinate plane all E_{y Q2} = E_Q2.

b) Field strength due to Charge Q1 is also given by

E_Q1 = k q/r²

The radius in this case is found by Pythagorean's Theorem

r = square root ( 11² * 17² ) = 20.25 cm

E_Q1 = (8.99 x 10⁹ N m² / C² ) (0.95 x 10^-9 C)/(0.2025 m)² = 208 N/C

To find the net Field Strength in the y-direction, we need the y-component of the field strength of from each of the individual charges. The electric field is all in the y-direction for Q2. So we only need to calculate the y-component of the Electric field strength from Q1.

The y-component of the field strength from Q1 is given by

E_yQ1 = E_Q1 sin Θ

where Θ is the angle of E1 relative to the x-axis.and is also the angle of r1 from the x-axis

so sin Θ = opposite/hypotenuse = 17/20.25

So E_yQ1 = (208 N/C) ( 17/20.25) = 175 N/C

The net field strength in the y-direction is the sum of the field strengths from the individual charges

E_{y net} = E_yQ1 + E_yQ2 = -2644 N/C + 175 N/C = -2569 N/C again the negative sign indicates the net field strength is in the negative y-direction.

c) To find the net field from both charges we add the contributions from each of the charges.

To do this we find the net field strength in the x-direction and the y-direction(done in part b). Since Q2 is directly below point P, all of the field strength from Q2 is in the y-direction and the x-component of the field strength from Q2 is zero

So, the net field strength in the x-direction is the x-component of the field strength from Q1. Which is given by

E_{x Q1} = E_{x net} = E_Q1 cos Θ = 208 N/C * (11/20.25) = 113 N/C

Now that we have the x-component and the y-component of the net field strength we can find the total net field strength using Pythagorean's Theorem.

E_{net =} Square Root ( (E_{x net})² + (E_{y net})² ) = Square Root ( (113 N/C)² + (-2569 N/C)² ) = 2571 N/C

Please

d) Let a = the angle of the net field strength from the x-axis. The tangent of angle a is given by

tan (a) = E_{y net} / E_{x net}

_{So a = arctan (} E_{y net} / E_{x net}) = arctan (-2569/113) = -87.5 degrees where the negative sign indicates that the direction of the net field strength is 87.5 degrees in the clockwise direction from the X-axis.

Best Regards,

Dean Creech