A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +12.0-ìC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm). (a) What is the change in the potential energy of the charge–field system? (b) Through what potential difference does the charge move?
Please help
Work= force dot displacement
= E*q dot displacement
= 250i*12 dot (.50j + .20i)
I assume you know how to do the dot product.
To calculate the change in potential energy of the charge-field system, we need to calculate the work done by the electric field on the charge as it moves from the origin to the given point (x, y) = (20.0 cm, 50.0 cm).
(a)
The work done by the electric field is given by the formula:
Work = Electric Field * Charge * Displacement * cosθ
In this case, the electric field (E) is given as 250 V/m, the charge (q) is 12.0 μC (which is equivalent to 12.0 x 10^(-6) C), and the displacement (d) is given as 0.20i + 0.50j.
The dot product between the force vector and the displacement vector is calculated as follows:
E * q * d = (250i * 12.0 x 10^(-6) C) * (0.20i + 0.50j)
To calculate the dot product, we multiply the corresponding components and take the sum of the products:
(250 * 12.0 x 10^(-6) * 0.20) + (250 * 12.0 x 0.50 x 10^(-6))
Simplifying the calculation further, we have:
(0.006) + (0.015)
= 0.021 J
Therefore, the change in potential energy of the charge-field system is 0.021 Joules.
(b)
The potential difference (V) through which the charge moves is equal to the work done (W) divided by the magnitude of the charge (|q|):
Potential Difference = Work / |q|
Substituting the values, we have:
Potential Difference = 0.021 J / 12.0 x 10^(-6) C
Simplifying the calculation further, we have:
Potential Difference = 1750 V
Therefore, the charge moves through a potential difference of 1750 Volts.