A charge Q= 2 C is distributed uniformly through out a bar of length L=2.5 m. The bar is placed horizontally in free space. A second charge q = 10−9C is placed along the line of the bar a distance d= 2m away in space, measured from the right end of the bar. What force is exerted on charge q by the charged bar?
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To find the force exerted on the charge q by the uniformly charged bar, we need to apply Coulomb's Law, which states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (Q1 and Q2) and inversely proportional to the square of the distance (r) between them:
\[ F = k_e \frac{|Q1 \cdot Q2|}{r^2} \]
However, since the charge Q is distributed along the length of the bar, we cannot treat it as a point charge. Instead, we need to consider the charge distribution and calculate the force exerted by each infinitesimally small segment of the bar on the charge q. To do this, we integrate the contribution of these segments along the length of the bar.
Let's define the following: - \( \lambda \) = Linear charge density of the bar (charge per unit length), which is Q/L. - dx = An infinitesimal segment's length of the bar. - \( dQ = \lambda \cdot dx \) = Charge of the infinitesimal segment. - x = Distance from the left end of the bar to the segment dx. - r = Distance from charge q to the segment dx. - \( k_e \) = Coulomb's constant, approximately \( 8.99 \times 10^9 \, N \cdot m^2/C^2 \).
First, let's find the linear charge density: \[ \lambda = \frac{Q}{L} = \frac{2 \, C}{2.5 \, m} = 0.8 \, C/m \]
Now, at any point (x) along the bar, the infinitesimal charge element \( dQ \) can be written as: \[ dQ = \lambda \cdot dx \]
The distance r from the charge q to an element dx on the bar can be written as: \[ r = (L - x) + d \] Here, (L - x) is the distance from the segment dx to the rightmost point of the bar, and d is the gap from the bar's right end to the charge q.
We plug these values into the formula for the force between point charges to find the infinitesimal force (dF) on the charge q caused by the element dx: \[ dF = k_e \frac{|dq \cdot q|}{r^2} \]
\[ dF = k_e \frac{|\lambda \cdot dx \cdot q|}{((L - x) + d)^2} \]
To find the total force, F, we integrate the expression for dF along the entire length of the bar from x = 0 to x = L: \[ F = \int_0^L k_e \frac{|\lambda \cdot q \cdot dx|}{((L - x) + d)^2} \]
\[ F = k_e \lambda q \int_0^L \frac{dx}{((L - x) + d)^2} \]
\[ F = k_e \lambda q \left[ - \frac{1}{(L - x) + d} \right]_0^L \]
\[ F = k_e \lambda q \left( - \frac{1}{d} + \frac{1}{(L + d)} \right) \]
\[ F = k_e \lambda q \left( \frac{1}{L + d} - \frac{1}{d} \right) \]
Substitute the given values \( \lambda = 0.8 \, C/m \), \( L = 2.5 \, m \), \( d = 2 \, m \), and \( q = 10^{-9} \, C \): \[ F = (8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (0.8 \, C/m) \cdot (10^{-9} \, C) \left( \frac{1}{2.5 \, m + 2 \, m} - \frac{1}{2 \, m} \right) \]
\[ F = (8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (0.8 \, C/m) \cdot (10^{-9} \, C) \left( \frac{1}{4.5 \, m} - \frac{1}{2 \, m} \right) \]
\[ F = (8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (0.8 \, C/m) \cdot (10^{-9} \, C) \left( \frac{2 - 4.5}{2 \cdot 4.5} \, m^{-1} \right) \]
\[ F = (8.99 \times 10^9 \, N \cdot m^2/C^2) \cdot (0.8 \, C/m) \cdot (10^{-9} \, C) \cdot (-\frac{2.5}{9}) \, m^{-1} \]
\[ F = -1.99 \times 10^{-2} \, N \]
The negative sign indicates that the force is attractive, meaning the charge q is pulled towards the charged bar. The calculated magnitude of the force is \( 1.99 \times 10^{-2} \, N \).
Extra: The concept of linear charge distribution is important when dealing with extended charges. Unlike point charges where charge is concentrated at a single point, linear charge distributions spread charge along a line. In this case, the bar has charge evenly distributed along its length, characterized by the linear charge density (λ), which represents how much charge is present per unit length of the bar.
Coulomb's Law, which we used to calculate the force, is a fundamental principle of electrostatics. This law enables us to calculate the electrostatic force between two point charges. However, for extended objects like the charged bar in this scenario, we cannot simply apply the formula directly. Instead, we must integrate over the entire charge distribution to account for the force contributed by each part of the charged object.
To integrate means to sum up the infinitesimal contributions from each differential segment of the charged object, considering the varying distances between these segments and the point charge they are interacting with. By doing this, we accurately calculate the force exerted by an extended charge distribution, such as a charged rod, on a point charge.