Please explain part B. I don\'t understand the direction of the vectors Adding e

Question

Please explain part B. I don't understand the direction of the vectors

Adding electric forces in one dimension Two +10nC charged particles are 2.0 cm apart on the .x-axis. What is the net force on a +1.0 nC charge midway between them? What is the net force if the charged particle on the right is replaced by a - 10 nC charge? prepare We proceed using the steps of Problem-Solving Strategy 20.1. We model the charged particles as point charges. The visual overview of FIGURE 20.16 establishes a coordinate system and shows the forces F_lon3 and F_2on3. Figure 20.16a shows a +10 nC charge on the right; Figure 20.16b shows a -10 nC charge. A visual overview of the forces for the two cases. Solve Electric forces are vectors, and the net force on q_3 is the vector sum F_net = F_1 on 3 + F_2 on 3. Charges q_1 and q_2 each exert a repulsive force on q_3, but these forces are equal in magnitude and opposite in direction. Consequently, F_net = 0. The situation changes if q_2 is negative, as in Figure 20.16b. In this case, the two forces are equal in magnitude but in the same direction, so F_net = 2F_1 on 3. The magnitude of the force is given by Coulomb's law. The force due to q_1 is F_1 on 3 = K|q1||q_3/r_13^2 = (9.0 times 10^9 N middot m^2/C^2)(10 times 10^-9 C)(1.0 times 10^-9 C)/(0.010 m)2 = 9.0 times 10^-4 N There is an equal force due to q_2, so the net force on the 1.0 nC charge is F_net = (1.8 times 10^-3 N, to die right). Assess This example illustrates the important idea that electric forces are vectors. An important part of assessing our answer is to see if it is "reasonable." In the second case, the net force on the charge is approximately 1 mN. Generally, charges of a few nC separated by a few cm experience forces in the range from a fraction of a mN to several mN. With this guideline, the answer appears to be reasonable.

Explanation / Answer

The answer to the first part is NOT 9.0x10 N. It is zero. The book answer is wrong.

This is because of symmetry; the charge in the middle is repelled equally in opposite directions. Go through the following steps for more detail:

Call the two equal 10nC charges 'A' and 'B'. Call the1nC charge 'C'. So the arrangement is:
A..C..B
Distance AC = BC = 1cm = 0.01m

C is repelled by A. This is a force of magnitude |F| acting on C pushing it to the right.
C is repelled by B. This is a force of magnitude |F| acting on C pushing it to the left.
The net force on C is therefore zero. This is true whatever the charges on A and B are - providing they are equal.

(If you work out the force between A and C. or B and C, you get:
F =kqq/d² = 9x10 x (10x10) x (1x10) / 0.01² = 9x10 N
but this value is not required for this part of the question.)

With B = -10nC

Force of A on C = kqq/d² = 9x10 x (10x10) x (1x10) / 0.01² = 9x10 N
The positive value indicates repulsion, so C is pushed to the right with a force of magnitude 9x10 N.

Force of B on C = kqq/d² = 9x10 x (-10x10) x (1x10) / 0.01² = -9x10 N
The negative value indicates attraction, so C is pulled to the right with a force of magnitude 9x10 N.

So C has 2 forces, each of magnitude 9x10 N, acting to the right.
Net force = 2 x 9x10 N = 1.8x10³ N.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---