A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f

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.

~Greetings! ^_^

Your inquiry: A plant produces flowers with a violet hue. The flowers possess a violet appearance due to their ability to absorb all light waves barring____ rays.

Your reply: A plant produces flowers with a violet hue. The flowers possess a violet appearance because they absorb all light rays except for violet rays.


Hope this is beneficial~

According to Newton's first law, an object remains at rest until an external force acts upon it, or an object in motion continues to move at a constant speed without accelerating.

Thus, x can solely represent a body coming to a halt. Accordingly, the last option is the most correct.

<span>Let Q be the charge, thus Q = -20.0 µC.</span>
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m, which represents the length of the rod
</span><span>To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
</span><span>E = K·Q/r²
</span>or<span>E = kQ/D(D+L), where k</span> is a constant equal to 8.99 x 10<span>9</span> N m
2/C2.<span>Consequently,[TAG_21]]E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>

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