A uniform electric field of magnitude 333 N/C is directed along the +y-axis. A 6.00 µC charge moves from the origin to the point (x, y) = (-10 cm, -37 cm).
A). What is the change in the potential energy associated with this charge?
B).Through what potential difference did the charge move?
The product of the E-field (in N/C) and the charge (in Coulombs) equals the force, which will be in the +y direction since the charge is positive.
Multiply the force by the distance moved IN THE FIELD (y) DIRECTION for the P.E change, but add a "minus" sign. The distance moved must be in meters.
The potential difference between the two points is E * (change in y)
Get to know these important concepts.
To find the change in potential energy associated with the charge, we can use the equation:
∆PE = q ∆V
where ∆PE is the change in potential energy, q is the charge, and ∆V is the potential difference.
A). Change in potential energy (∆PE):
Since the electric field is uniform, the potential difference ∆V is equal to the negative of the dot product between the electric field and the displacement vector (∆r):
∆V = -E ∆r
The electric field E is given as 333 N/C and the displacement vector ∆r is the distance between the initial position (origin) and the final position (-10 cm, -37 cm). To calculate the displacement vector, we can use the Pythagorean theorem:
∆r = sqrt((∆x)^2 + (∆y)^2)
where ∆x and ∆y are the differences in the x-axis and y-axis positions, respectively.
∆x = -10 cm = -0.1 m
∆y = -37 cm = -0.37 m
∆r = sqrt((-0.1 m)^2 + (-0.37 m)^2) = 0.383 m
Now, substitute these values into the formula for ∆V:
∆V = -(333 N/C)(0.383 m) = -127.939 N·m/C
Since the charge q is given as 6.00 µC = 6.00 x 10^(-6) C, we can now find the change in potential energy ∆PE:
∆PE = (6.00 x 10^(-6) C)(-127.939 N·m/C) = -0.767634 J
Therefore, the change in potential energy associated with this charge is approximately -0.767634 J.
B). Potential difference (∆V):
The potential difference (∆V) is equal to the negative work done by the electric field on the charge q:
∆V = -W
The work done (W) by the electric field is given by the equation:
W = F ∆d
where F is the magnitude of the force experienced by the charge and ∆d is the magnitude of the displacement.
The force experienced by the charge in an electric field is given by:
F = qE
where q is the charge and E is the magnitude of the electric field.
Substituting these values into the equation for work done, we get:
W = (qE) ∆d
In this case, q = 6.00 µC = 6.00 x 10^(-6) C, E = 333 N/C, and ∆d = 0.383 m.
∆V = -W = -(6.00 x 10^(-6) C)(333 N/C)(0.383 m) = -0.766074 J
Therefore, the charge moved through a potential difference of approximately -0.766074 J. Note that ∆V is negative because the charge is moving against the electric field.