1. The diagram shows two points in an electric field: point 1 is at (x 1 , y 1 )

Question

1.
The diagram shows two points in an electric field: point 1 is at (x1, y1) = (7, 4), and point 2
is at (x2, y2) = (12, 9), in metres.

The electric field is constant, with a magnitude of 71 V/m, and is directed parallel to the +x axis. The potential at point 1 is 1100 V. Calculate the potential at point 2. Hint: To calculate the potential between points 1 and 2 you will need to choose a path. You will get the same answer no matter what path you choose.

2.
Calculate the work required to move a negative charge of q = ?608 µC from point 1 to point 2.

Explanation / Answer

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

1) Calculate the potential at point 2
we went from x = 7 to x = 12, this is a difference of 5m. Field is 71 V/m, so every time we go a meter we lose 71 V
1100.0V - 5m*71 V/m = 745 V

2) Calculate the work required to move a negative charge of Q=-608 C from point 1 to point 2

assuming a negative charge of Q=-608 C
Volt = work/coulomb
Voltage difference*(-608E-6C)=work
(1100 - 745)*(608 E-6C) = 0.215 J

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