Question

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. After the charge has moved to point B, 0.500 m to the right, it has kinetic energy5.00×10−7 J .A. If the electric potential at point A is +30.0 V, what is the electric potential at point B?B. What is the magnitude of the electric field?C. What is the direction of the electric field?a. from point B to point Ab. from point A to point B c. perpendicular to the line AB

Answer 1

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$= 0) to the point B at distance l = 0.500m where its kinetic energy is (  $K_{B}$= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA,$K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

                                   $K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)                                          

Where $K_{A}$= 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential.  So, equation (1) will be in the form :

                                  0+qVA=$K_{B}$ +qVB                      (Divide by q)

                                         VA=$K_{B}$ /q + VB                  (solve for VB)

                                         VB=VA- $K_{B}$/q .......................................(2)

We get the relation between VB, VA and $K_{B}$, now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

                                         VB=VA- $K_{B}$/q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

                                              =80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

                                   VA-VB =$\int\limits^1_0 {E} \, dl$...................................(a)

                                               $=E\int\limits^1_0 {} \, dl$

                                   VA-VB=E$l$                      (solve for E)

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

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

                                            E= VA-VB/$l$

                                              =-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge