PHYSICS QUESTION PLEASE HELP!!!?
Two point charges, Q1 = 3.2 μC and Q2 = -1.3 μC , are placed on the x axis. Suppose that Q2 is placed at the origin, and Q1 is placed at the coordinate x1 = − 4.0 cm (Figure 1).
A) At what point(s) along the x axis is the electric field zero? Determine the x-coordinate(s) of the point(s).
B) At what point(s) along the x axis is the potential zero? Determine the x-coordinate(s) of the point(s). (!!!!!2 coordinates please!!!!)
- NCSLv 71 month agoFavorite Answer
A) field E = vector Σ kQ / d²
Since the two charges have opposite sign, there is ONE point where the field is zero, and it lies OUTSIDE of the two charges and close to the one with smaller magnitude. So we need the point where
k*3.2µC/(4.0cm + x)² = k*1.3µC/x²
3.2x² = 1.3(4 + x)²
for x in cm.
This quadratic has roots at
x = -1.6 cm ← ignore negative root
and x = 7.0 cm ◄
B) potential U = algebraic Σ kQ / d
There can't be a point to the left of Q1 where this is true.
If there is a point between the two charges, then
k*3.2µC / (4.0cm - x) = k*1.3µC / x
3.2x = 1.3(4 - x)
for x in cm.
This has a solution at x = 1.16 cm, which puts the point at
X = -1.16 cm ≈ -1.2 cm
to two significant digits.
If there is a point to the right of Q2, then
k*3.2µC / (4.0cm + x) = k*1.3µC / x
3.2x = 1.3(4.0 + x)
This has a solution at
x = 2.7 cm
for a total of two such points.
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- az_lenderLv 72 months ago
I have just seen your comments on my previous answer. Sorry to say, I am mystified. Glad they agree that my answer to (A) was correct and that the one answer I gave for (B) is also correct, but you say they allege that part (B) has a 2nd answer. I can't understand that. Any point to the left of both charges will be more affected by the Q1 than by the Q2; and any point between the two charges will have potentials that add together rather than tending to cancel (?) Hope another user can help you...
- Anonymous2 months ago
find it yourself