Three charges lie on a line: A = +5.0 µC at x = 0, B = -2.0 µC at x = 0.20 m, and C = +3.0 µC at x…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Three charges lie on a line: A $= +5.0\ \mu\text{C}$ at $x = 0$ , B $= -2.0\ \mu\text{C}$ at $x = 0.20$ m, and C $= +3.0\ \mu\text{C}$ at $x = 0.50$ m. Find the force of A on B, and the force of C on B. Then find the net force on B. Finally find the net force on C, and say which way it points.

Need a reminder? The key ideaShow

The net force on a charge is the vector sum of the forces from all the others. On a line, add with signs. In a plane, add components and use $F = \sqrt{F_x^2 + F_y^2}$ . Between two like charges the force on a third is zero where $\dfrac{x}{d - x} = \sqrt{\dfrac{q_1}{q_2}}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find each force on its own. Choose $+x$ as positive, and add with signs.

Step 1. A on B. The distance is $0.20$ m. $F = \dfrac{kq_Aq_B}{r^2}$ . The top is $9.0 \times 10^9 \times 5.0 \times 10^{-6} \times 2.0 \times 10^{-6}\ \text{N m}^2 = 0.090\ \text{N m}^2$ and the bottom is $(0.20\ \text{m})^2 = 0.040\ \text{m}^2$ . So $F = 2.25\ \text{N}$ . It is an attraction, towards A, so $-2.25$ N.

Step 2. C on B. The distance is $0.30$ m. $F = \dfrac{kq_Cq_B}{r^2}$ . The top is $9.0 \times 10^9 \times 3.0 \times 10^{-6} \times 2.0 \times 10^{-6}\ \text{N m}^2 = 0.054\ \text{N m}^2$ and the bottom is $(0.30\ \text{m})^2 = 0.090\ \text{m}^2$ . So $F = 0.60\ \text{N}$ . It is an attraction, towards C, so $+0.60$ N.

Step 3. The net force on B. $-2.25\ \text{N} + 0.60\ \text{N} = -1.65\ \text{N}$ . The size is $1.65$ N, towards A.

Step 4. The net force on C. A is $0.50$ m from C. For A on C, the top is $9.0 \times 10^9 \times 5.0 \times 10^{-6} \times 3.0 \times 10^{-6}\ \text{N m}^2 = 0.135\ \text{N m}^2$ and the bottom is $(0.50\ \text{m})^2 = 0.25\ \text{m}^2$ . So $F_{AC} = 0.54\ \text{N}$ , a push to $+x$ . B is $0.30$ m from C. For B on C, the top is $0.054\ \text{N m}^2$ and the bottom is $0.090\ \text{m}^2$ . So B pulls C towards B with $0.60\ \text{N}$ , to $-x$ . So $F = 0.54\ \text{N} - 0.60\ \text{N} = -0.06\ \text{N}$ .

Answer: $1.65$ N towards A on B, and $0.06$ N towards B on C.

Check. The forces on B are of different sizes, so there is a net force. The two forces on C are nearly equal, so the net force on C is small.

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