The diagram shows two points in an electric field: point 1 is at (x 1 , y 1 ) =
ID: 1320673 • Letter: T
Question
The diagram shows two points in an electric field: point 1 is at (x1, y1) = (9, 4), and point 2
is at (x2, y2) = (12, 9), in metres.
The electric field is constant, with a magnitude of 68 V/m, and is directed parallel to the +x axis. The potential at point 1 is 1000 V. Calculate the potential at point 2.
Calculate the work required to move a negative charge of q = ?532 C from point 1 to point 2.
The diagram shows two points in an electric field: point 1 is at (x1, y1) = (9, 4), and point 2 is at (x2, y2) = (12, 9), in metres. The electric field is constant, with a magnitude of 68 V/m, and is directed parallel to the +x axis. The potential at point 1 is 1000 V. Calculate the potential at point 2. Calculate the work required to move a negative charge of q = ?532 ½C from point 1 to point 2.Explanation / Answer
E = (V2-V1)/d...
given that V1 = 1000 V
E = 68 V/m
V2-V1 = E*d = 68*sqrt((12-9)^2+(9-4)^2) = 396.5 V
V2 = 396.5+1000 = 1396.5 V
B) W = q*(V2-V1) = -532*10^-6*396.5 = 0.21 J