The drawing shows a uniform electric field that points in the negative y direction; the magnitude of the field is 4540 N/C. Determine the electric potential difference the following points

VB − VA between points A and B
VC − VB between points B and C
VA − VC between points C and A
the picture is a triangle,the x axis is 6cm,the y axis is 8cm, and the hypothenus is 10

User Icon for bobpursley bobpursley answered
13 years ago

Well, I have no idea of the picture. But E is a vector, and displacement is a vector.

Voltage= abs E * abs d which is the magnitude of E times the magnitude of distance from the points * cosine Theta where Theta is the angle between E and the displacement vector.

User Icon for OSUstudent OSUstudent answered
13 years ago

I had this question too- think abstractly and it's pretty easy.

In general: E = delta V/ delta s

Vb-Va = 0 (the points are at the same location in terms of the equipotential surface)

Vc-Vb = E* delta s (or the distance between the points)

so 4540*0.08m = 363 V

Va-Vc: a is the same distance from c as b when thinking in therms of the surface (0.08m) however, your answer will have the opposite sign because you are traveling in the opposite direction of the electric field.

so 4540*0.08m = -363 V

User Icon for Anonymous Anonymous answered
13 years ago

thank you

User Icon for Clown Bot Clown Bot answered
1 year ago

Well, considering the electric field is pointing in the negative y direction, it seems like it's having a negative effect on the vibes around here. Let's calculate the electric potential difference!

Now, let's start with points A and B. Since the electric potential difference is given by VB - VA, we need to find the potential at each point.

Since the electric potential difference is equal to the negative of the work done in moving a positive charge from one point to another, we have:

VB - VA = - ∫(E dot dl)

However, in this case, we have a uniform electric field, which means the electric field is constant everywhere. Therefore, we can simply use:

VB - VA = - Ed

Now, since the magnitude of the electric field is given as 4540 N/C and the distance between points A and B is 8 cm (since the y axis is 8 cm), we can plug in the values:

VB - VA = - (4540 N/C)(0.08 m)

VB - VA = - 363.2 N\

Now, let's move on to points B and C. Again, we will use the same formula, but this time the distance between B and C is the base of the triangle, which is given as 6 cm:

VC - VB = - (4540 N/C)(0.06 m)

VC - VB = - 272.4 N

Finally, let's tackle points C and A. Since we're going from C to A, the distance between C and A will be the hypotenuse of the triangle, which is conveniently given as 10 cm:

VA - VC = - (4540 N/C)(0.10 m)

VA - VC = - 454 N

So, the electric potential differences are:
VB - VA = -363.2 N
VC - VB = -272.4 N
VA - VC = -454 N

And there you have it! Now, let's hope these electric potential differences don't start any shocking conversations!

User Icon for Explain Bot Explain Bot answered
1 year ago

To determine the electric potential difference between the given points, we need to use the formula ΔV = -Ed, where ΔV is the potential difference, E is the electric field magnitude, and d is the distance or displacement between the points.

Let's calculate the distances between the points first using the given dimensions of the triangle. In this case, we have a right-angled triangle with the x-axis measuring 6 cm, the y-axis measuring 8 cm, and the hypotenuse measuring 10 cm.

Using the Pythagorean theorem, we can calculate the distance between the points:

AB: The distance between points A and B is equivalent to the length of the y-axis, which is 8 cm.

BC: The distance between points B and C is the difference between the length of the hypotenuse (10 cm) and the length of the y-axis (8 cm). Thus, BC = 10 cm - 8 cm = 2 cm.

CA: The distance between points C and A can be found by applying the Pythagorean theorem to the given triangle, considering the x-axis and y-axis lengths. Hence, CA is equal to the square root of [(y-axis)^2 + (x-axis)^2] = √(8 cm)^2 + (6 cm)^2 = √(64 cm^2 + 36 cm^2) = √100 cm^2 = 10 cm.

Now that we have the distances between the points, we can compute the electric potential differences:

VB - VA between points A and B:
ΔVAB = -(E)(dAB) = -(4540 N/C)(8 cm) = -36320 N·cm/C.

VC - VB between points B and C:
ΔVBC = -(E)(dBC) = -(4540 N/C)(2 cm) = -9080 N·cm/C.

VA - VC between points C and A:
ΔVCA = -(E)(dCA) = -(4540 N/C)(10 cm) = -45400 N·cm/C.

Therefore, the electric potential differences are:
VB - VA = -36320 N·cm/C
VC - VB = -9080 N·cm/C
VA - VC = -45400 N·cm/C.