(a)Two charges are placed on the x-axis: one is placed at x = 3 m and the other is at x = -3 m. The magnitude of both charges is 2.3 µC but the blue one (at x = -3 m) is positive while the red one (at x = +3 m) is negative.
What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?
(b) Now the positive and negative charge switch places. The magnitude of the charges is still 2.3 µC where the blue one (now at x = +3 m) is positive and the red one (now at x = -3 m) is negative.
What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?
(c) Now both charges (still at x = -3 m and x = +3 m) are positive. The magnitude of both charges is still 2.3 µC.
What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?
(d) Finally, both charges (still at x = -3 m and x = +3 m) are negative. The magnitude of both charges is still 2.3 µC.
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Answer:
A. 15x +14y = -36
Step-by-step explanation:
Since we are given two points, we can start with the 2-point form of the equation for a line.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (6 -(-9))/(-8 -6)(x -6) +(-9)
y = 15/-14(x -6) -9
Multiplying by -14, we have ...
-14y = 15x -90 +126
Adding 14y-36 to both sides gives ...
-36 = 15x +14y . . . . matches choice A
The standard-form equation is ...
15x +14y = -36
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<em>Additional comments</em>
It can be easier to start with the form ...
(Δy)x -(Δx)y = (Δy)x1 -(Δx)y1 . . . . . where Δx = x2-x1 and Δy = y2-y1
This gives ...
(6+9)x -(-8-6)y = 15(6) +14(-9)
15x +14y = -36 . . . simplified
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You can also start with the slope-intercept form or the point-slope form, if you're more familiar with those. The result will be the same. I find it handy to be familiar with a number of different forms of the equation for a line.
C. 96 students will not prefer math or language arts
Answer:
What are you solving for?
Step-by-step explanation: