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

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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:

kicyunya [3.2K] · Answer 1 · 2026-04-02T08:01:54+03:00

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 ( $K_{A}$ = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 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, $K_{B}$ 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:

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

Where $K_{A}$ = 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= $K_{B}$ +qVB (Dividing by q)

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

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

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

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

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

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

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

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

E= VA-VB/ $l$

=-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.