A point charge q1 = -1.6 C is at the origin, and a second point charge q2 = +4.6

Question

A point charge q1 = -1.6 C is at the origin, and a second point charge q2 = +4.6 C is at the point x = 1.0 m, y = 2.0 m.
(a) Find the x and y coordinates of the position at which an electron would be in equilibrium.
x ______ = 1 m
y ______ = 2 m

(b) Find the x and y coordinates of the position at which a proton would be in equilibrium.
x______ = 3 m
y______ = 4 m

(c) Suppose the charges q1 and q2 were both doubled. How would your answers change?
The electron would be in equilibrium at the following point.
x ______ = 5 m
y_______ = 6 m

The proton would be in equilibrium at the following point.
x_____ = 7 m
y______ = 8 m

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Explanation / Answer

q1 = -1.6 C at O(0, 0), q2 = +4.6 C at A(1.0, 2.0) m
(a) Find the x and y coordinates of the position at which an electron would be in equilibrium.

The electron must be at P(x, y) on the line connecting O and A. Note q1 and q2 have different signs, P must not be between O and A. Also note |q1| < q2, P must be at 3 quadrant.

The line connecting O and A has the equation y = 2x.

The distance AO = d = 5 m, the distance OP = (x2 + y2) = [x2 + (2x)2] = x5 = xd

For equilibrium, |q1|/d2 = q2/(d + xd)2

(1 + x)2 = q2/|q1|

1 + x = (q2/|q1|)

x = (q2/|q1|) - 1 = 0.7

y = 2x = 1.4

answer:

x = -0.7 m

y = -1.4 m

(b) Find the x and y coordinates of the position at which a proton would be in equilibrium.
same answer as (a)

x = -0.7 m
y = -1.4 m

(c) Suppose the charges q1 and q2 were both doubled. How would your answers change?
The electron would be in equilibrium at the following point.

x = (q2/|q1|) - 1 , so when q2 and q1 were both doubled, x would have no change.

x = -0.7 m
y = -1.4 m

The proton would be in equilibrium at the following point.
x = -0.7 m
y = -1.4 m

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