three charges are located at the vertices of a triangle, as shown. +5 c dot, +10 c dot, +4 c dot with vectors w (left), x (up - right), y (right), z (down - right). which vector best represents the net force acting on the +4 c charge in the diagram? options: w, x, y, z (each with a circle).

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# Brief Explanations:
1. First, recall Coulomb's law: like charges repel each other. So the +5 C and +10 C charges will repel the +4 C charge.
 - The force from +5 C on +4 C: Since both are positive, the force (let's call it $ \vec{F}_{5 - 4} $) is directed away from +5 C (towards the general direction opposite to +5 C relative to +4 C).
 - The force from +10 C on +4 C: Both positive, so the force ($ \vec{F}_{10 - 4} $) is directed away from +10 C (towards the general direction opposite to +10 C relative to +4 C).
2. Now, analyze the magnitudes: The magnitude of the electric force is given by $ F = k\frac{q_1q_2}{r^2} $. The +10 C charge has a larger magnitude than +5 C. Assuming the distances from +4 C to +5 C and +4 C to +10 C are such that the force from +10 C ($ F_{10 - 4} $) is stronger than the force from +5 C ($ F_{5 - 4} $) (since $ q_2 = 10 $ C is larger than $ q_1 = 5 $ C, and if distances are comparable, the larger charge gives a stronger force).
3. The net force is the vector sum of $ \vec{F}_{5 - 4} $ and $ \vec{F}_{10 - 4} $. Let's consider the directions:
 - The force from +5 C on +4 C: Let's say the +5 C is at the top - left, +10 C at the bottom - left, and +4 C in the middle - right. So the force from +5 C on +4 C is a repulsion, so direction is towards the right - downish (but need to consider the triangle). Wait, actually, in the diagram, the +5 C is above the +10 C, and both are to the left of +4 C. So the force from +5 C on +4 C is a repulsion, so direction is away from +5 C: so from +4 C, towards the right - down (since +5 C is up - left). The force from +10 C on +4 C is a repulsion, away from +10 C: so from +4 C, towards the right - up (since +10 C is down - left). But since +10 C has a larger charge, its force is stronger. So the vector sum of a force from +5 C (let's say with some direction) and a stronger force from +10 C (with another direction) will result in a net force. Wait, maybe a better way: both +5 C and +10 C are on the left of +4 C, repelling it. So the force from +5 C is to the right (since repulsion, so +4 C is pushed away from +5 C, which is on the left - up, so the force from +5 C is towards right - down? No, wait, the position: +5 C is at (let's imagine coordinates) (0, 5), +10 C at (0, 0), +4 C at (5, 0). Then the force from +5 C on +4 C: the vector from +5 C to +4 C is (5, - 5), so the force (repulsive) is in the same direction (since like charges repel, so the force on +4 C is away from +5 C, so direction from +4 C to away from +5 C, which is (5, - 5) direction? Wait, no: the force on +4 C due to +5 C is $ \vec{F} = k\frac{q_1q_2}{r^2}\hat{r} $, where $ \hat{r} $ is the unit vector from +5 C to +4 C. So if +5 C is at (x1,y1) and +4 C at (x2,y2), then $ \hat{r}=(x2 - x1,y2 - y1)/r $. So the force on +4 C is in the direction away from +5 C, so same as $ \hat{r} $. Similarly for +10 C.
 - Now, the +10 C charge is larger, so the force from +10 C on +4 C is stronger than from +5 C. So the net force should be a vector that is the sum of two repulsive forces: one from +5 C (smaller magnitude) and one from +10 C (larger magnitude). Let's look at the vectors: W is left, X is up - right, Y is right, Z is down - right.
 - The force from +5 C: since +5 C is above +10 C, the force from +5 C on +4 C will have an upward component (because +5 C is above) and a rightward component (since it's to the left of +4 C). The force from +10 C on +4 C will have a downward component (since +10 C is below) and a rightward component (since it's to the left of +4 C). But since +10 C has a larger charge, the downward component from +10 C's force will be stronger than the upward component from +5 C's force? Wait, no, if +5 C is at the top, +10 C at the bottom, and +4 C to the right. Then the force from +5 C on +4 C: direction is from +4 C away from +5 C, so towards the bottom - right (because +5 C is top - left, so away is bottom - right). The force from +10 C on +4 C: direction is from +4 C away from +10 C, so towards the top - right (because +10 C is bottom - left, so away is top - right). Wait, now I'm confused. Maybe the diagram is such that +5 C and +10 C are on the left, with +5 C above +10 C, and +4 C on the right. So the two forces on +4 C are:
   - From +5 C: repulsion, so direction is to the right - down (since +5 C is up - left, so pushing +4 C down - right).
   - From +10 C: repulsion, so direction is to the right - up (since +10 C is down - left, so pushing +4 C up - right).
   - Now, the magnitude of the force from +10 C is larger (because $ q = 10 $ C > $ q = 5 $ C, and if distances are similar, $ F\propto q $). So the upward component from +10 C's force is stronger than the downward component from +5 C's force. So the net force will have a rightward component (since both forces have rightward components) and an upward component (since +10 C's upward component is stronger than +5 C's downward component). Looking at the vectors: X is up - right, Z is down - right, Y is right, W is left. So the net force should be in the direction of X? Wait, no, wait: maybe the +10 C is more massive? No, charge, not mass. Wait, Coulomb's law: $ F = k\frac{q_1q_2}{r^2} $. So if the distance from +4 C to +10 C is $ r_1 $ and to +5 C is $ r_2 $. But in the triangle, maybe the distance to +10 C is shorter? Wait, the diagram shows a triangle with +5 C and +10 C connected vertically, and both connected to +4 C. So the distance from +4 C to +10 C might be shorter than to +5 C? Or maybe the same? Wait, the problem is about the net force vector. Let's think again: like charges repel. So +5 C repels +4 C (force direction: from +4 C away from +5 C), +10 C repels +4 C (force direction: from +4 C away from +10 C). The +10 C has a larger charge, so its repulsive force is stronger. So the vector from +10 C's force is more significant. So the net force should be a combination of the two repulsive forces. If +5 C is above +10 C, then the force from +5 C is down - right, and from +10 C is up - right. Since +10 C's force is stronger, the up - right component is stronger, so the net force is up - right, which is vector X? Wait, no, maybe I got the directions wrong. Wait, the +4 C is in the middle, +5 C is top - left, +10 C is bottom - left. So the force on +4 C from +5 C: the line connecting +5 C and +4 C, so the force is along that line, away from +5 C (so from +4 C, towards the direction opposite to +5 C, i.e., down - right). The force from +10 C: along the line connecting +10 C and +4 C, away from +10 C (so from +4 C, towards up - right). Now, the magnitude of the force from +10 C is $ F_{10}=k\frac{4\times10}{r_{10}^2} $, and from +5 C is $ F_5 = k\frac{4\times5}{r_5^2} $. If the distance $ r_{10} $ is equal to $ r_5 $ (since it's a triangle, maybe isoceles), then $ F_{10}=2F_5 $. So the force from +10 C is twice as strong as from +5 C. So the vector for $ F_{10} $ is up - right, magnitude 2F, and $ F_5 $ is down - right, magnitude F. So the net force is (2F up - right)+(F down - right)= F up - right. So the net force is in the up - right direction, which is vector X? Wait, but let's check the options: W is left, X is up - right, Y is right, Z is down - right. So the net force should be X? Wait, no, maybe I messed up the direction of the force. Wait, the force on +4 C due to +5 C: the +5 C is a positive charge, so it repels +4 C, so the force on +4 C is directed away from +5 C. So if +5 C is at (0, 5) and +4 C at (5, 0), the vector from +5 C to +4 C is (5, - 5), so the force on +4 C is in the direction (5, - 5) (away from +5 C). The force due to +10 C (at (0, 0)) on +4 C (at (5, 0)): the vector from +10 C to +4 C is (5, 0), so the force on +4 C is in the direction (5, 0) (away from +10 C)? Wait, no, if +10 C is at (0, 0) and +4 C at (5, 0), then the distance is 5 units along the x - axis. So the force on +4 C due to +10 C is in the positive x - direction (since repulsion). The force on +4 C due to +5 C (at (0, 5)): the distance is $ \sqrt{(5 - 0)^2+(0 - 5)^2}=\sqrt{50} $, and the force direction is from +4 C away from +5 C, so the vector is (5 - 0, 0 - 5)=(5, - 5), unit vector is $ (\frac{5}{\sqrt{50}},\frac{- 5}{\sqrt{50}})=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}) $. So the force from +5 C is $ F_5 = k\frac{4\times5}{50}=k\frac{20}{50}=k\frac{2}{5} $, direction $ (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}) $. The force from +10 C is $ F_{10}=k\frac{4\times10}{25}=k\frac{40}{25}=k\frac{8}{5} $, direction (1, 0) (since along x - axis). Now, let's find the components:
 - $ F_{10} $ components: $ (k\frac{8}{5},0) $
 - $ F_5 $ components: $ (k\frac{2}{5}\times\frac{1}{\sqrt{2}},k\frac{2}{5}\times(-\frac{1}{\sqrt{2}}))=(k\frac{\sqrt{2}}{5},-k\frac{\sqrt{2}}{5}) $
 - Net force components: $ F_x = k\frac{8}{5}+k\frac{\sqrt{2}}{5}\approx k(\frac{8 + 1.414}{5})\approx k\frac{9.414}{5} $, $ F_y = 0 - k\frac{\sqrt{2}}{5}\approx - k\frac{1.414}{5} $
 Wait, this gives a net force with positive x and negative y (down - right), which would be vector Z. But this contradicts the earlier thought. What's wrong here? Ah, maybe the position of +10 C is not at (0, 0) but at (0, - 5), and +5 C at (0, 5), +4 C at (5, 0). Then the force from +10 C (at (0, - 5)) on +4 C (at (5, 0)): vector from +10 C to +4 C is (5, 5), so force direction is (5, 5) (away from +10 C), unit vector $ (\frac{5}{\sqrt{50}},\frac{5}{\sqrt{50}})=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) $. Force magnitude $ F_{10}=k\frac{4\times10}{50}=k\frac{40}{50}=k\frac{4}{5} $. Force from +5 C (at (0, 5)) on +4 C (at (5, 0)): vector (5, - 5), unit vector $ (\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}) $, magnitude $ F_5 = k\frac{4\times5}{50}=k\frac{20}{50}=k\frac{2}{5} $. Now, net force components:
 - $ F_x = k\frac{4}{5}\times\frac{1}{\sqrt{2}}+k\frac{2}{5}\times\frac{1}{\sqrt{2}}=k\frac{6}{5\sqrt{2}}\approx k\frac{6}{7.071}\approx k\frac{0.848}{1} $
 - $ F_y = k\frac{4}{5}\times\frac{1}{\sqrt{2}}-k\frac{2}{5}\times\frac{1}{\sqrt{2}}=k\frac{2}{5\sqrt{2}}\approx k\frac{2}{7.071}\approx k\frac{0.282}{1} $
 So net force is in the up - right direction (positive x and positive y), which is vector X. Ah, so the key was the position of +10 C. If +10 C is below +5 C (so at (0, - 5)) and +4 C at (5, 0), then the force from +10 C has an upward component (since it's below +4 C, repelling it upwards) and the force from +5 C has a downward component (since it's above +4 C, repelling it downwards). Since +10 C has a larger charge, its upward component is stronger than +5 C's downward component, resulting in a net upward - right force, which is vector X. Wait, but in the diagram, the +10 C is below +5 C (connected by a vertical line), and +4 C is to the right. So the two forces on +4 C are:
 - From +5 C (top - left): repulsion, so force direction is down - right (away from +5 C).
 - From +10 C (bottom - left): repulsion, so force direction is up - right (away from +10 C).
 Since +10 C has a larger charge, the up - right force from +10 C is stronger than the down - right force from +5 C. So the net force is the vector sum: (up - right force)+(down - right force)= (stronger up - right)+(weaker down - right)= net force with right and up components, which is vector X. Wait, but in my first coordinate system, I had +10 C at (0, 0), which was wrong. The correct position is +10 C below +5 C, so at (0, - h) and +5 C at (0, h), +4 C at (d, 0). Then the force from +10 C is up - right, force from +5 C is down - right. Since +10 C has more charge, the up - right force is stronger, so net force is up - right (vector X).
# Answer:
X

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