Question

If the separation of the plates is reduced, but the potential difference across them remains constant, which of the following statements must be true? 1\. The electric field increases. 2\. The magnetic field increases. 3\. The capacitance increases. 1\. I only 2\. I and II only 3\. I and III only 4\. II ant III only

Short answer

When the separation of the plates is reduced and the potential difference remains constant, both the electric field and the capacitance increase. Therefore, the correct option is 3. I and III only.

Step-by-step solution

Statement 1: The electric field increases.

First, let's discuss how the electric field between the plates of a capacitor can be affected when reducing the separation. The electric field (E) between the plates of a capacitor can be calculated using the relation: \(E = \frac{V}{d}\) where V is the potential difference across the capacitor and d is the separation between the plates. Since the potential difference (V) remains constant, if the separation (d) is reduced, the electric field (E) will increase. So, statement 1 is true.

Statement 2: The magnetic field increases.

The magnetic field between the plates of a capacitor is not affected by the separation between the plates. Instead, it depends on the current passing through the device. Since there is no information given about any change in the current, we cannot conclude that the magnetic field increases. Statement 2 may not be true.

Statement 3: The capacitance increases.

The capacitance (C) of a capacitor is directly proportional to the surface area of the plates (A) and inversely proportional to the separation between the plates (d). The formula for capacitance is given by: \(C = \frac{ε_0 A}{d}\) where ε_0 is the vacuum permittivity. In this case, since we are reducing the separation (d) while keeping the surface area of the plates (A) constant, the capacitance (C) will increase. Statement 3 is true. Based on our analysis, statements 1 and 3 are true. So, the correct answer is: 3. I and III only