Does the position of a point influence the electric potential between the origin and that point?

Question

A uniform electric field of a magnitude of 8 x 10^3 N/C points in the positive x direction. Find the change in electrical potential between the origin and the points (a) (0, 7.0m), (b) (7.0m, 0), and (c) (7.0m, 7.0m).

(a) (0, 7.0m) is ___ V
(b) (7.0m, 0) is ___ V
(c) (7.0m, 7.0m) is ___ V

Answer 1

answer is definetly zero but i can explain u why it is zero
as u know electric field is responsible for force applied on any charge, and when we move a charge against this electric field say a+ve charge is brought near b positive charge u have to do a work on this charge to take it closer to the b+ve charge this work get stored in the body as energy and it is known as change in potential
if u think why u have uto do a work it is because electric field is not constant

now consider ur question electric field is constnt means at all the point take 10 km 10m all will have same electric field from origin point charge hence think deeply, u don't have to do any work on bringing it near or far hence no energy change hence no potential diffrence

this is basic of potential energy and potential diffrence if u think like this ur electostat will become strong

think practically about potential

Answer 2

The direction of the field is in the x direction.

Uniform electric field means the field is constant at points along a given direction.

The potential difference between two points is the work done in bringing a unit positive charge from one point to another point against the field of force.

a)
(0, 7).

The point is in the y axis at a distance of 7 from the origin.

The line joining the origin and the point is perpendicular to the x direction or the direction of the field.

Therefore, no work is done in bringing the unit positive charge from the point to the origin.

The potential difference between the two points is zero. Both points are in the same potential.


b)
(7,0)

The point in the x axis is at a distance of 7 from the origin.

Work has to be done to bring a unit positive charge from this point to the origin.

P.D = 8 x 10^3 x 7 = 56 x 10^3 V.

c)
(7,7)

The point is along at a distance of 7 from x-axis and 7 from y axis.
Along the y direction no work is done. But along x axis work is done.

P.D = 8 x 10^3 x 7 = 56 x 10^3 V.

Answer 3

hey only in case all people else seems up this question, billrussel42 and m6 are incorrect as a manner to get the voltage distinction between A and B you shuttle from A to B interior the path of the present and sum up the effective factors and losses in skill yet you will first could calculate the flair during the left, middle, and appropriate wires. Designate randomly that I1 is during the left cord and travels up, that I2 is during the middle cord and travels down, and that I3 is during the main suitable cord and travels down. Then use the loop regulation and junction rule to get those equations: I1 = I2+I3, -8.5I1-4-2I2=0, and -12-4I3+2I2=0. fixing those equations supplies I1=-.814A, I2=a million.5A, and I3=-2.27A. this means that our I1 and I3 have been interior the incorrect guidelines, and we now understand that I1 is going down and I3 is going up. Then we calculate the voltage distinction between A and B via first taking place the left cord: +4V-8.5(I1) = 4-8.5(.814) = -2.ninety two V.

Answer 4

The answer to a is zero, and b and c are the same. The result will be 56E3 volts. (I am assuming that newtons per coulomb and volts per meter are equivalent; it has been a while since I did this sort of thing.)