This is for my practice exam. I really want to be able to study from this exampl

Question

This is for my practice exam. I really want to be able to study from this example so if someone could answer it while writing out ALL of the steps you took i would REALLY appreciate it. If you could show me the steps with the final answer that would be perfect.

1. Two point charges are placed as follows:
[q1= +0.25 x 109C] at (1,0) and [q2= -2 x 10-9C] at (1, 3)

a.) What is the electric field (MAGNITUDE and DIRECTION) at the origin? Begin by drawing a diagram (please show the diagram!!) that shows schematically the placement of the charges and the direction of the electric field vectors at the origin due to each charge; use k = 9 x 109 Nm2/C2

b.) What is the electric potential at the origin?

c.) If we assume the electric potential is 0 at infinity (that is, very far away), how much work has to be done to bring a charge of +1 C from very far away (infinity) to the origin?

Explanation / Answer

charges q1 = 0.25*10^-9 C             q2 = -2*10^-9 C electric field at origin due to ( 1,0 ) is E = Kq1 / x^ 2 where K = coulomb's constant = 9*10^9 Nm^ 2/ c^ 2          x = distance of origin from q1 = 1 m plug the values we get E = 2.25 N / C it is towards q1 to origin x-component of electric field at origin due to ( 1,3 ) is E ' = Kq2 / x^ 2          x = distance of origin from q2 = 1 m plug the values we get E' = 18 N / C it is towards origin to point q2 y-component of electric field at origin due to ( 1,3 ) is E " = Kq2 / y^ 2          y = distance of origin from q2 = 3 m plug the values we get E " = 6 N / C it is towards positive y -axis net x-component of field E1 = E ' -E = 15.75 N / C it is along positive x-axis net electric field = E1 i + E " j                      E2 = 15.75 i + 6 j magnitude of E2 is = [15.75^2 + 6^ 2]                               = 284 N / C Let E2 makes an angle with x -axis then tan = 6 / 15.75     = 20.85 degrees (b). Potential at origin due to q1 is V1 = Kq1 / x                                                           = 2.25 volt potential at origin due to q2 is V2 = Kq2 / r where r = distance of origin from q2 = [1^ 2 + (3)^ 2 ]                                                          = 2m So, V2 = 9 volt So, potential at origin V = V1 + V2 = 11.25 volt x-component of electric field at origin due to ( 1,3 ) is E ' = Kq2 / x^ 2          x = distance of origin from q2 = 1 m plug the values we get E' = 18 N / C it is towards origin to point q2 y-component of electric field at origin due to ( 1,3 ) is E " = Kq2 / y^ 2          y = distance of origin from q2 = 3 m plug the values we get E " = 6 N / C it is towards positive y -axis net x-component of field E1 = E ' -E = 15.75 N / C it is along positive x-axis net electric field = E1 i + E " j                      E2 = 15.75 i + 6 j magnitude of E2 is = [15.75^2 + 6^ 2]                               = 284 N / C Let E2 makes an angle with x -axis then tan = 6 / 15.75     = 20.85 degrees (b). Potential at origin due to q1 is V1 = Kq1 / x                                                           = 2.25 volt potential at origin due to q2 is V2 = Kq2 / r where r = distance of origin from q2 = [1^ 2 + (3)^ 2 ]                                                          = 2m So, V2 = 9 volt So, potential at origin V = V1 + V2 = 11.25 volt y-component of electric field at origin due to ( 1,3 ) is E " = Kq2 / y^ 2          y = distance of origin from q2 = 3 m plug the values we get E " = 6 N / C it is towards positive y -axis net x-component of field E1 = E ' -E = 15.75 N / C it is along positive x-axis net electric field = E1 i + E " j                      E2 = 15.75 i + 6 j magnitude of E2 is = [15.75^2 + 6^ 2]                               = 284 N / C Let E2 makes an angle with x -axis then tan = 6 / 15.75     = 20.85 degrees (b). Potential at origin due to q1 is V1 = Kq1 / x                                                           = 2.25 volt potential at origin due to q2 is V2 = Kq2 / r where r = distance of origin from q2 = [1^ 2 + (3)^ 2 ]                                                          = 2m So, V2 = 9 volt So, potential at origin V = V1 + V2 = 11.25 volt y-component of electric field at origin due to ( 1,3 ) is E " = Kq2 / y^ 2          y = distance of origin from q2 = 3 m plug the values we get E " = 6 N / C it is towards positive y -axis net x-component of field E1 = E ' -E = 15.75 N / C it is along positive x-axis net electric field = E1 i + E " j                      E2 = 15.75 i + 6 j magnitude of E2 is = [15.75^2 + 6^ 2]                               = 284 N / C Let E2 makes an angle with x -axis then tan = 6 / 15.75     = 20.85 degrees (b). Potential at origin due to q1 is V1 = Kq1 / x                                                           = 2.25 volt potential at origin due to q2 is V2 = Kq2 / r where r = distance of origin from q2 = [1^ 2 + (3)^ 2 ]                                                          = 2m So, V2 = 9 volt So, potential at origin V = V1 + V2 = 11.25 volt

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