S13Phys2BaLec21C

# If you have more than two charges present you must

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multiple charges present. If you have more than two charges present you must use vector superposition. This means you calculate each force separately and then add them together under the rules of vector addition. You have to take both magnitude and direction into account.

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Superposition Example Calculate the net electric force on the rightmost particle (q 3 ) in the figure below due to the other two charges. Answer First, you must define a coordinate system. Let’s say that the rightmost particle is located at x = 0 and that the left direction is positive. 0.30m 0.20m q 1 =–3.0 μ C q 2 =+5.0 μ C q 3 =–4.0 μ C +x
Superposition Answer Next, let’s list the quantities that we know: q 1 = –3.0x10 –6 Coul q 2 = +5.0x10 –6 Coul q 3 = –4.0x10 –6 Coul r 23 = 0.20m r 13 = 0.50m Let’s calculate each force separately. First, let’s start the force between particles 2 and 3.

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Superposition Answer Next, let’s calculate the force between particles 1 and 3. What about direction? Since particles 2 and 3 are oppositely charged, they will be attractive => F 23 is in +x direction. Since particles 1 and 3 are like charged, they will be repulsive => F 13 is in the –x direction.
Superposition Answer Next, we add the two forces (via superposition). Σ F elec = F 23 + F 13 = 4.5N + –0.43N = +4.1N Where the positive sign indicates that the force points to the left (since we chose left as +). The answer is the net electric force will have a magnitude of 4.1N and point to the left. Note: The middle charge q 2 in no way blocks the effect of q 1 . If q 2 did not exist q 1 would still contribute the same amount.

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Superposition Example In the figure below, particles 1 and 2 are fixed in place on the x-axis at a separation of 8.00cm. Their charges are q 1 =+e and q 2 =-27e. Particle 3 with charge q 3 =+4e is to be placed on the line between particles 1 and 2, so that they produce a net electrostatic force on it. At what coordinate should
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