A uniform electric field ofmagnitude 384 V/m is directed in the positive x-direction. Suppose a 16 ìC charge moves from the origin to point A at the coordinates, (20 cm, 60 cm). What is the absolute value of the change in
potential from the origin to point A?
Answer in units of V
To calculate the change in potential, we need to know the formula for potential in an electric field and the distance between the origin and point A.
The potential in an electric field is given by the equation:
V = E * d
Where:
V is the potential
E is the electric field magnitude
d is the distance
In this case, the electric field magnitude (E) is given as 384 V/m.
To find the distance (d) between the origin and point A, we can use the Pythagorean theorem as the distance is given in terms of coordinates:
d = √(x^2 + y^2)
In this case, x = 20 cm = 0.2 m and y = 60 cm = 0.6 m.
Plugging the values into the equation:
d = √((0.2 m)^2 + (0.6 m)^2)
= √(0.04 m^2 + 0.36 m^2)
= √0.4 m^2
= 0.6325 m
Now we can substitute the values of E and d into the formula to calculate the potential change:
V = E * d
= 384 V/m * 0.6325 m
= 243.84 V
Therefore, the absolute value of the change in potential from the origin to point A is 243.84 V.