Problem 113 - Master Resource Book in JEE Main Physics

Question

The force on each plate of parallel plate capacitor has a magnitude equal to \(\frac{1}{2} Q E\), where \(Q\) is the charge on the capacitor and \(E\) is the magnitude of electric field between the plates. Then (a) \(\frac{E}{2}\) contributes to the force against which the plates are moved (b) \(\frac{E}{3}\) contributes to the force against which the plates are moved (c) \(E\) contributes force against which the plates are moved (d) None of the above

Step-by-Step Solution

Answer

(a) \( \frac{E}{2} \) contributes to the force against which the plates are moved.

Step 1: Understanding the Concept of Electric Field in a Capacitor

In a parallel plate capacitor, the electric field (E) between the plates is uniform and is given as \( E = \frac{V}{d} \), where \( V \) is the potential difference and \( d \) is the separation between the plates. This field acts uniformly across the plate surfaces.

Step 2: Analyzing the Force on Each Plate

The problem states that the force on each plate is \( \frac{1}{2} Q E \). This implies that the interaction of the charges with the electric field between the plates results in this force. Generally, the electric field created by one plate exerts a force on the opposite plate.

Step 3: Contribution of Electric Field to Plate Movement

Given the force on a plate is due to the interaction between the plates, it is necessary to consider components of this electric field. Since only half of the total field created by one plate interacts with the other plate, it's effectively \( \frac{E}{2} \). The remaining field contributes to the force exerted against the internal field of the same plate.

Step 4: Choosing the Correct Answer

Since it is the field \( \frac{E}{2} \) from one plate that contributes to the force of attraction or repulsion that needs to be overcome to move a plate, the correct choice is (a) \( \frac{E}{2} \) contributes to the force against which the plates are moved.

Key Concepts

Electric Field

The electric field, often denoted as \(E\), is a fundamental concept in electromagnetism. In the context of parallel plate capacitors, the electric field is uniform and exists between the plates. This uniformity means that the electric field strength is the same at every point between the plates. The formula to calculate the electric field \(E\) in a capacitor is given by \(E = \frac{V}{d}\), where \(V\) is the potential difference across the plates, and \(d\) is the distance separating them.

In simple terms, the electric field is the force per unit charge that a positive test charge would experience between the plates. This field arises because one plate has a positive charge, while the other possesses an equal negative charge, leading to a straight path for the electric force acting on charged particles in the space between. Understanding this can help visualize how charges interact within capacitors and why their spatial arrangement affects how capacitors function.

Forces in Capacitors

When discussing forces in capacitors, it is important to understand how the electric field interacts with the charges on the plates. The force exerted on each plate is due to the electric field generated by the charge on the opposite plate. The problem specifically notes that this force has a magnitude of \(\frac{1}{2}Q E\).

This force arises because the electric field from one plate attracts or repels the charges on the other plate. However, it's not the entire electric field \(E\) that does this. Only half of it, \(\frac{E}{2}\), effectively contributes to moving the plates. This is due to the fact that each plate creates its own electric field that acts in opposition to the field created by the other plate.

In practical terms, this means that the moving or separating of plates requires overcoming only half the field's effect from the opposite plate. Understanding this subtle detail clarifies how the forces inside capacitors are balanced and what determines their magnitude.

Charge Interactions

Charge interactions in a parallel plate capacitor are pivotal in determining how the device functions. Each plate in the capacitor becomes uniformly charged, with one holding a positive charge and the other an equal negative charge. These charges interact through the electric field formed between them, creating the forces that we have discussed.

The interesting part about charge interactions is how they lead to the attraction and repulsion forces. The positively charged plate attracts negative charges, while it repels positives, and vice versa for the negatively charged plate. This interaction is what creates the potential energy stored within the capacitor.

When thinking about how this works, it is essential to realize that the charges don't just stay on the plates; they interact through the electric field, and their distribution across the plate surfaces impacts the efficiency and capacity of the capacitor. By understanding charge interactions, one can gain insights into how capacitors store energy and how adjustments in plate separation and charge amount influence their performance.