PS-4-Prep-2-SS2013 Part A A Q,--14 (Charge is placed on the x-axis at x =-0.1 m.
ID: 1533766 • Letter: P
Question
PS-4-Prep-2-SS2013 Part A A Q,--14 (Charge is placed on the x-axis at x =-0.1 m. A QR-14 pC charge is placed at x +0.1 m. (for all answers below assume that the unit vector points toward positive x, and ý points towards positive y). What is the x-component of the electric field at x 0 m and y Express your answer using three significant figures. 0.1 m? N/C Submit My Answers Give Up Part B What is the y-component of the electric field at x 0 m and y 0.1 m? y N Submit My Answers Give Up Part C What is the x-component of the electric field at x 0 m and y--01 m? Submit My Answers Give UpExplanation / Answer
The first determine the direction of the E fields at (0,0.1)
The field at (0,0.1) due to the positive charge points at 135 degrees.
The field at (0,0.1) due to the negative charge points at 225 degrees.
This is because the E field is OUT of positive charges and IN to negative charges.
The distance squared from each charge to (0,0.1) is 0.1^2 + 0.1^2 = 0.02 m
Since both charges have the same magnitude and are the same distance from (0,0.1),
their fields have the same intensity. The magnitude of the E fields at (0,0.1) is
k*14*10^-6/0.02
8.99*10^9 * 14*10^-6/0.02
= 6293000 N/C
The net E field at (0,0.1) is the sum of the two vector E fields from the two charges.
Sum the horizontal components of the fields at (0,0.1)
6293000*cos135 + 6293000*cos225 = -3956839.965 N/C
Sum the vertical components
6293000*sin135 + 6293000*sin225 = -5296982.925 N/C
Therefore the net E field at (0,0.1) is
(a) 3956839.965 N/C at 180 degrees
(b) 5296982.925 N/C
(c) 3956839.965 N/C
(d) 5296982.925 N/C
(c) and (d) have the same answers as (a) and (b) because of the symmetry. (0,0.1) is the same distance from both charges and field directions are the same.
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