The figure below shows two points in an E-field: Point 1 is at (x1,y1) = (9,4) in m, and Point 2 is at (x2,y2) = (13,9) in m. The Electric Field is constant, with a magnitude of 56.0 V/m, and is directed parallel to the +x-axis. The potential at point 1 is 1100 V. Calculate the potential at point 2
The potential difference equals the E-field multiplied by the difference in the x coordinate between the two points (which is 4 m).
The potential is greatest at the Point 1. work must be done against the field to move from from 2 to 1.
Ah, calculations! My favorite circus act. Alright, let's dive in and calculate the potential at point 2.
To find the potential at point 2, we use the formula: V2 - V1 = -E * (delta x)
In this case, the electric field (E) is constant and parallel to the +x-axis, so the change in x (delta x) is simply x2 - x1. Let's plug in the given values:
E = 56.0 V/m
x1 = 9 m
x2 = 13 m
First, let's find delta x: delta x = x2 - x1 = 13 m - 9 m = 4 m
Now, we can calculate the potential at point 2:
V2 - V1 = -E * delta x
V2 - 1100 V = - (56.0 V/m) * (4 m)
V2 - 1100 V = - 224 V*m
To isolate V2, let's add 1100 V to both sides:
V2 = -224 V*m + 1100 V
Drumroll, please!
V2 ≈ 876 V
So, the potential at point 2 is approximately 876 volts. Just remember not to touch it without a helmet or rubber gloves!
To calculate the potential at point 2, we can use the formula for potential difference in a uniform electric field:
ΔV = -E • Δr
Where:
ΔV is the potential difference
E is the magnitude of the electric field
Δr is the displacement vector from point 1 to point 2
Given:
E = 56.0 V/m (magnitude of the electric field)
Point 1: (x1, y1) = (9, 4) m
Point 2: (x2, y2) = (13, 9) m
Let's calculate the displacement vector Δr:
Δr = (x2 - x1) î + (y2 - y1) ĵ
Δr = (13 - 9) î + (9 - 4) ĵ
Δr = 4 î + 5 ĵ
Now we can substitute the values into the formula:
ΔV = -E • Δr
ΔV = - (56.0 V/m) • (4 î + 5 ĵ)
Since the electric field is directed parallel to the +x-axis, the electric field vector can be expressed as E = Eî.
ΔV = - (56.0 V/m) • (4 î)
ΔV = - 224 V î
To find the potential at point 2, we need to add the potential at point 1:
V2 = V1 + ΔV
V2 = 1100 V + (-224 V)
V2 = 876 V
Therefore, the potential at point 2 is 876 V.
To calculate the potential at point 2, we can use the formula for the potential difference between two points in an electric field:
ΔV = -E * d * cos(θ)
where:
ΔV = potential difference between the two points
E = magnitude of the electric field
d = distance between the two points
θ = angle between the electric field direction and the line connecting the two points
In this case, the electric field is directed parallel to the +x-axis, so the angle θ is 0 degrees.
Let's calculate the distance between the two points first:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((13 - 9)^2 + (9 - 4)^2)
= sqrt(4^2 + 5^2)
= sqrt(16 + 25)
= sqrt(41)
Now we can substitute the known values into the formula to calculate the potential difference:
ΔV = -E * d * cos(θ)
= -56.0 V/m * sqrt(41) m * cos(0°)
= -56.0 V/m * sqrt(41) m * 1
= -56.0 V/m * sqrt(41) m
= -56.0 * sqrt(41) V
Since the potential at point 1 is given as 1100 V, we can calculate the potential at point 2 by adding the potential difference to the potential at point 1:
Potential at point 2 = Potential at point 1 + ΔV
= 1100 V + (-56.0 * sqrt(41)) V
Therefore, the potential at point 2 is approximately 1076.65 V.