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Brrunno [24]
3 years ago
7

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. Aft

er 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
Physics
1 answer:
sergeinik [125]3 years ago
5 0

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:

<u>Given :</u>

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.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

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

<u> Solution </u>

(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=El                      (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

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