The figure below shows two points in an electric field. Point 1 is at (X_1, Y_1)

Question

The figure below shows two points in an electric field. Point 1 is at (X_1, Y_1) = (3, 4), and point 2 is at (X_2, Y_2) = (12, 9). The electric field is constant with a magnitude of 80.3 V/m, and is directed parallel to the +X-axis. The potential at point 1 is 1000.0 V. Calculate the work required to move a negative charge of Q = -562.0 µC from point 1 to point 2. A: 1.33 times 10^-1 B: 1.93 times 10^-1 C: 2.80 times 10^-1 D: 4.06 times 10^-1 E: 5.89 times 10^-1 F: 8.54 times 10^-1 G: 1.24 H: 1.80

Explanation / Answer

Electric field is horizontal, so no work to move a charge vertically

1) we went from x = 3 to x = 12,

this is a difference of 9m.

Field is 80.3 V/m, so every time we go a meter we lose 80.3 V
1000.0V - 9m*80.3V/m = 277.3V

2) Volt = work/coulomb
Voltage difference*(-562.0E-6C) = work
(277.3 - 1000)*(-562*10^-6C) = +0.4061 J = 4.061*10^-1 J (option D)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---