A thin semicircular rod has a total charge +2 uniformly distributed along it. A negative point charge-Q is placed as shown. A test charge +q is placed at point C. (Point C is equidistant from -Qand from all points on the rod.)
Let F, and F represent the force on the test charge due to the point charge and the rod respectively.
a. Is the magnitude of F, greater than, less than, or equal to the magnitude of Fa? Explain how you can tell.
The Correct Answer and Explanation is :
Let’s analyze the situation step by step.
The test charge ( +q ) is placed at point C, which is equidistant from the negative point charge ( -Q ) and from all points along the uniformly charged semicircular rod. We need to compare the magnitudes of the forces ( F_1 ) and ( F_2 ), where:
- ( F_1 ) is the force on the test charge due to the point charge ( -Q ),
- ( F_2 ) is the force on the test charge due to the uniformly charged rod.
Step 1: Force due to the point charge ( -Q ) ( ( F_1 ) )
The force exerted on the test charge ( +q ) by a point charge ( -Q ) is given by Coulomb’s Law:
[
F_1 = k_e \frac{|Q| \cdot |q|}{r^2}
]
where:
- ( k_e ) is Coulomb’s constant,
- ( Q ) is the charge of the point charge ( -Q ),
- ( q ) is the charge of the test charge ( +q ),
- ( r ) is the distance between the point charge ( -Q ) and the test charge ( +q ).
Since the test charge is equidistant from the point charge and the rod, the magnitude of ( F_1 ) is dependent on this distance ( r ).
Step 2: Force due to the uniformly charged rod ( ( F_2 ) )
The force on the test charge due to the uniformly charged semicircular rod is more complex to calculate, as it involves the integration of forces from each infinitesimal element of charge along the rod. However, we know that due to the symmetry of the setup, the rod’s total effect on the test charge at point C will result in a force that is primarily directed radially outward from the center of the rod. The magnitude of this force depends on the total charge on the rod and the distance from the test charge to the rod.
Comparison of ( F_1 ) and ( F_2 )
Given the configuration:
- The point charge ( -Q ) is a single point, which results in a relatively stronger force due to the inverse square law (since force decreases rapidly with distance squared).
- The rod, on the other hand, spreads out its charge over a curved length. Even though the rod has a uniform charge distribution, the net effect of all the infinitesimal elements of charge will lead to a weaker force compared to the point charge, because the rod’s charge is spread out over a semicircular arc.
Thus, ( F_1 ) (the force due to the point charge) is greater than ( F_2 ) (the force due to the uniformly charged rod). The point charge exerts a stronger force because the force from the rod is diluted over a curved distribution, while the point charge exerts its force in a direct, concentrated manner.
Now, I’ll generate the image to visually represent the situation described.
Here is the diagram illustrating the setup you described, with the forces ( F_1 ) and ( F_2 ) acting on the test charge. As we discussed, ( F_1 ), the force from the point charge, is greater than ( F_2 ), the force from the rod. This is due to the concentrated nature of the point charge’s force compared to the distributed effect of the rod’s charge.
