Question

Eight \(1.00-\mu C\) charges are arrayed along the \(y\) -axis located every \(2.00 \mathrm{~cm}\) starting at \(y=0\) and extending to \(y=14.0 \mathrm{~cm} .\) Find the force on the charge at \(y=4.00 \mathrm{~cm} .\)

Short answer

Question: Determine the net force acting on a positive charge (1.00 -$\mu$C) placed at y=4.00 cm along the y-axis, given the presence of seven other equal positive charges (1.00 -$\mu$C) located at y= 0, 2, 6, 8, 10, 12, and 14 cm along the y-axis. Answer: After following the steps provided in the solution to find the net force acting on the target charge, we find that the net force is approximately -0.470 N (downward).

Step-by-step solution

Identify target charge and other charges

The target charge is located at \(y=4.00 \mathrm{~cm}\), and there are seven other charges located along the y-axis as given in the problem. We need to calculate the force exerted by each of those seven charges on our target charge.

Sign of the forces

The charge of all the particles is positive (\(1.00 -\mu C\)). As a result, the forces between our target charge and the other charges will be repulsive. The force exerted by a charge below the target charge will be in the downward direction, and the force exerted by a charge above the target charge will be in the upward direction.

Calculating distances from target charge

Measure the distance between the target charge and every other charge, in centimeters: - At \(y=0.00 \mathrm{~cm}\): \(d_1 = 4.00 \mathrm{~cm}\) - At \(y=2.00 \mathrm{~cm}\): \(d_2 = 2.00 \mathrm{~cm}\) - At \(y=6.00 \mathrm{~cm}\): \(d_3 = 2.00 \mathrm{~cm}\) - At \(y=8.00 \mathrm{~cm}\): \(d_4 = 4.00 \mathrm{~cm}\) - At \(y=10.0 \mathrm{~cm}\): \(d_5 = 6.00 \mathrm{~cm}\) - At \(y=12.0 \mathrm{~cm}\): \(d_6 = 8.00 \mathrm{~cm}\) - At \(y=14.0 \mathrm{~cm}\): \(d_7 = 10.0 \mathrm{~cm}\)

Convert distances and charges

Convert distances from centimeters to meters, and charges from \(-\mu C\) to Coulombs: - \(d_1 = 0.0400\,\mathrm{m}\), \(d_2 = 0.0200\,\mathrm{m}\), ..., \(d_7 = 0.100\,\mathrm{m}\) - \(q = 1.00 \times 10^{-6}\,\mathrm{C}\)

Calculate forces exerted by other charges

Use Coulomb's Law to calculate the force (\(F_i\)) exerted by each charge (\(q_i\)) on the target charge (\(q_T\)) located at \(y=4.00 \mathrm{~cm}\): $$ F_i = k \frac{ q_i q_T }{d_i^2}$$ Compute the forces exerted by each charge and consider their direction (+ or -): - \(F_1 = 8.99 \times 10^9 \cdot \frac{(1 \times 10^{-6})(1 \times 10^{-6})}{0.0400^2} = 0.0562\,\mathrm{N}\) (downward) - \(F_2 = 8.99 \times 10^9 \cdot \frac{(1 \times 10^{-6})(1 \times 10^{-6})}{0.0200^2} = 0.224\,\mathrm{N}\) (downward) - ... And so on for other forces.

Calculate the net force

Sum up the forces on the target charge considering their directions: $$ F_{\mathrm{net}} = F_1 + F_2 - F_3 - F_4 - F_5 - F_6 - F_7 $$ After calculating the net force, you will have the force acting on the target charge at \(y=4.00 \mathrm{~cm}\).