A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height

Response:

a) 80 V

b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.

Clarification:

Given:

An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( = 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.

Required:

(a) We seek to find the electric potential VB

(b) We need to compute the magnitude and orientation of the electric field E.

Solution

(a) Utilizing the given values for VA, and q, we derive a relationship among the three parameters and VB to compute VB.

At points A and B, the charge moves from A to B due to the electric field. The mechanical energy of the object remains conserved throughout this journey, allowing us to apply eq(1) in this context:

                                   .........................................(1)                                          

Where = 0, and the potential energy U of the charge is defined as U = q V

In this equation, V represents the electric potential. Thus, equation (1) can be expressed as:

                                  0+qVA= +qVB                    (Dividing by q)

                                         VA= /q + VB                  (Restructuring for VB)

                                         VB=VA- /q.......................................(2)

We now have the relation between VB, VA, and , allowing us to substitute our values for VA, , and q into equation (2) to obtain VB

                                         VB=VA- /q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x 10^-9)

                                              =80 V

(b) After calculating VB, we may use equation a to derive the electric field E affecting the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between these points

                                   VA-VB =...................................(a)

                                               

                                   VA-VB=E                      (Restructuring for E)

                                            E= VA-VB/..................................(3)

Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E

                                            E= VA-VB/

                                              =-100 N/C

The electric field's magnitude equals 100 N/C and it directs from point B to point A towards the negative charge.

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