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

Answer: The net force acting on the charge at y = 4.00 cm is obtained by calculating the forces exerted by the other charges, particularly the charges at y = 0.00 cm (q₁) and y = 2.00 cm (q₂), and then subtracting the force exerted by q₁ from that exerted by q₂, as they act in opposite directions. The net force on the charge at y = 4.00 cm is given by F = F₂₃ - F₁₃.

Step-by-step solution

Charge Distribution and Data

Here, we have eight \(1.00-\mu C\) charges along the y-axis. They are arranged at equal distances of \(2.00 \, cm\). We need to find the force on the charge at \(y = 4.00 \, cm\). Let's denote this charge as \(q_3\).

Coulomb's Law

According to Coulomb's Law, the force between two point charges is given by $$F = k\frac{|q_1q_2|}{r^2},$$ where \(k = 8.99 \times 10^9 \, Nm^2/C^2\) is the Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.

Individual Forces

Compute the forces exerted by all other charges on charge \(q_3\). Let's denote the force between charges \(q_i\) and \(q_j\) as \(F_{ij}\). Due to the symmetry of distribution of charges and because the forces are along y-axis, the forces exerted by charges at equal distances above and below charge \(q_3\) will cancel out. Thus, we only need to find the net force that \(q_1\) and \(q_2\) exert on \(q_3\).

Force Due to \(q_1\)

Calculate the force exerted by charge \(q_1\) on charge \(q_3\) (\(F_{13}\)). Here, the distance between \(q_1\) and \(q_3\), \(r_{13} = 4.00 \, cm = 0.0400 \, m\) and \(q_1 = q_3 = 1.00 \, \mu C\). $$F_{13} = k\frac{|q_1q_3|}{r_{13}^2} = 8.99 \times 10^9 \frac{(1.00 \times 10^{-6})^2}{(0.040)^2}.$$

Force Due to \(q_2\)

Calculate the force exerted by charge \(q_2\) on charge \(q_3\) (\(F_{23}\)). Here, the distance between \(q_2\) and \(q_3\), \(r_{23} = 2.00 \, cm = 0.0200 \, m\) and \(q_2 = q_3 = 1.00 \, \mu C\). $$F_{23} = k\frac{|q_2q_3|}{r_{23}^2} = 8.99 \times 10^9 \frac{(1.00 \times 10^{-6})^2}{(0.020)^2}.$$

The Net Force

Now find the net force on charge \(q_3\). Since both \(F_{13}\) and \(F_{23}\) are on opposite directions, we need to subtract \(F_{13}\) from \(F_{23}\), that is $$F = F_{23} - F_{13}.$$

Key concepts

Electric Charge Distribution

Understanding electric charge distribution is crucial when calculating forces in electrostatics. Charges can be distributed in different ways, such as uniformly over a surface, along a line, or spaced out at intervals. The problem, in this case, involves a linear array of equal charges spaced at regular intervals along the y-axis.

Charge distribution impacts the electrostatic force experienced by any individual charge within the system. The symmetry in charge distribution plays a pivotal role in simplifying force calculations. For example, charges equally spaced on opposite sides of a central charge can produce forces that cancel each other out due to their symmetry. This is why in our exercise we observed that only the nearest neighbors on either side of the central charge contribute to the net force on that charge.

In essence, understanding the pattern in which charges are arrayed helps us determine the forces acting between the charges, making it easier to calculate the net electrostatic force on any given charge.

Electrostatic Force Calculation

The task of calculating electrostatic force between charges is made manageable by using Coulomb's Law, which gives us a straightforward formula. The electrostatic force (\( F \)) between two point charges is proportional to the product of the charges (\( q_1 \) and \( q_2 \)) and inversely proportional to the square of the distance (\( r \)) between them.

More succinctly, the formula is:\[\begin{equation}F = k\frac{{|q_1q_2|}}{{r^2}},\text{where } k = 8.99 \times 10^9 \, Nm^2/C^2\text{ is the Coulomb's constant.}\text{end equation}
Solving our problem, we calculate the forces between the charge at \( y=4.00 \text{ cm} \) (\( q_3 \)) and its nearest charges (\( q_1 \) and \( q_2 \)). Due to the linear distribution, we only consider the forces along the y-axis ensuring we are considering vector quantities with both magnitude and direction.

To compute the net force on \( q_3 \), we determine the force from each neighbor and account for their directions – forces from charges below \( q_3 \) act upwards along the y-axis, while those above act downwards. Then by subtracting these individual forces, considering their directions, we determine the net force experienced by \( q_3 \).

Principle of Superposition in Electrostatics

The principle of superposition in electrostatics is critical for solving complex force interaction problems. This principle states that the total force acting on a charge is the vector sum of all the individual forces exerted on that charge by other charges within the system.

In practice, this means we calculate the force between each pair of charges separately and then combine these forces to determine the net force on any charge. When charges exert forces along the same line, as in our example, the vector addition simplifies to algebraic addition with careful attention to the direction of each force.

For our exercise, we leveraged the principle of superposition by calculating the forces from charges \( q_1 \) and \( q_2 \) on \( q_3 \) separately and then subtracting one from the other to find the net force. It's essential to recognize that if there were any other charges not symmetrical to \( q_3 \) or on different axes, their forces would also need to be included in our superposition to truly capture the total force on \( q_3 \).