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Question

Three small negatively charged metal spheres in vacuum are fixed on a horizontal straight line, the x-axis. One (-15.5 µC) is at the origin, another (+7.00 µC) is at x = 2.0 m, and the third (-50.0 µC) is 1.0 m beyond that at x = 3.0 m. Compute the magnitude of the net electric force on the last sphere due to the other two.


pls show me the steps you took to get your answer and what you get..thanks!!

Answer 1

The problem is as straightforward as it can be. But, first, you must make your mind up as to whether the spheres are all negatively charged or not. Indeed, the second sphere, charged with +7 µC, has not a negative charge!

You can solve this problem either by figuring out the resultant force on the -50 µC charge, directly; or, you may find the resultant electric field at the point at which this charge rests, then computing total force on the charge. Either way, you'll use the Superposition Principle to get the result. Briefly, this means that you can calculate the force (or field) the first charge exerts on the third as if the second charge weren't there. Conversely, you figure out the force which the second charge applies on the third, as if the first were inexistent.

Fortunately, should the second charge also be negative, you'll only need to change a sign to arrive at the right answer. Although it could seem unlikely, the field approach is somewhat simpler from a computational standpoint, and will be used it here. Field E at x = 3 is given by

E₃= kq₁/r₁₃² + kq₂/r₂₃² = k(q₁/r₁₃² + q₂/r₂₃²),

where k = 8.99755 E9 N m²/C², r₁₃ means the distance between points at which spheres 1 and 3 are located (that is, 3 m); likewise, r₂₃ stands for the distance between spheres 2 and 3 (1 m). Force on a charge q placed at a point where the electric field is E, is given by F = qE, so

F = qE = -50 E-6 × 8.9755 E9(-15.5 E-6 / 3² + 7 E-6 / 1²) = -2.375 N.

This should be construed as an attractive force, in the negative direction of the x-axis. Should the second sphere be charged with a negative charge of the same magnitude, just swap the positive sign in front of the “7” in the above formula to get 3.925 N, a repulsive force, in the positive direction of the x-axis.

Clearly, the force exerted by the charge on the second sphere, which is nearer to the third sphere, overrides the other force, even when its charge is smaller. This is so because force is inversely proportional to the SQUARE of the distance separating the spheres.

Answer 2

1. -15.5 µC at x=0
2. +7.00 µC is at x = 2.0 m
3. -50.0 µC is at x=3.0 m

Ft= -F1,3 + F2,3

F= kQ1Q2/R^2

Ft=k( - (15.5E-6) (50.0 E-6)/4 +(7.0 E-6)(50.0E-6)/9)
k=8.988E+9 N m^2 C^2
Ft=8.988E+9 (-1.9375E-10 + .389E-10)= - 1.4N
the force will be repulsive directed towards positive x-axis

Answer 3

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