Figure 15 shows a long, insulating, massless rod of length L, pivoted at its cen

Question

Figure 15 shows a long, insulating, massless rod of length L, pivoted at its center and balanced with a weight Watt a distance x from the left end. At the left and right ends or the rod are attached positive charges q and 2q, respectively. A distance h directly beneath each of these charges is a fixed positive charge Q. (a) Find the distance x for the position of the weight when the rod is balanced. (b) What value should h have so that the rod exerts no vertical force on the bearing when balanced? Neglect the interaction between charges at the opposite ends of the rod.

Explanation / Answer

Hi, From the figure it can be seen that there will be repulsive force between the two charges on left side and also between the two charges on right side. The force of repulsion on the left side will be (k * q^2 / h^2) and that on the right side will be (k * 2 * q^2 / h^2). Hence there will be an additional repulsive force(in upward direction on 2q) of (k * q^2 / h^2) on the right side. Let the weight W added at a distance of x from the left most end such that it balances this additional force (as weight acts downwards). So now we have two forces, one upwards at a distance of L/2 from the pivot and another downwards at a distance of (x - L/2) from the pivot which are balanced. => (k * q^2 / h^2) * L/2 = W * (x - L/2)   => x = [(k * q^2 / h^2) * L/ (2 * W)] + L/2. b) From the above equation, we can also derive the value of 'h' for which when W is placed at x, as h = (k * q^2 * L) / [2 * W * (x - L/2)] Both h and x are interdependent. Hope this helps you. Hi, From the figure it can be seen that there will be repulsive force between the two charges on left side and also between the two charges on right side. The force of repulsion on the left side will be (k * q^2 / h^2) and that on the right side will be (k * 2 * q^2 / h^2). Hence there will be an additional repulsive force(in upward direction on 2q) of (k * q^2 / h^2) on the right side. Let the weight W added at a distance of x from the left most end such that it balances this additional force (as weight acts downwards). So now we have two forces, one upwards at a distance of L/2 from the pivot and another downwards at a distance of (x - L/2) from the pivot which are balanced. => (k * q^2 / h^2) * L/2 = W * (x - L/2)   => x = [(k * q^2 / h^2) * L/ (2 * W)] + L/2. b) From the above equation, we can also derive the value of 'h' for which when W is placed at x, as h = (k * q^2 * L) / [2 * W * (x - L/2)] Both h and x are interdependent. Hope this helps you. Both h and x are interdependent. Hope this helps you.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.