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
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
thank you
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!
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.
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.