Question

force acting upon a charged particle kept between the plates of a charged condenser is f. if one of the plates of the condenser is removed, force acting on the same particle will become 1) zero 2) f/2 3) f 4) 2f

Explanation:

Step1: Recall Electric Field in Capacitor

In a charged capacitor (condenser), the electric field between the plates is \( E=\frac{\sigma}{\epsilon_0} \) (for a parallel - plate capacitor, where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space). The force on a charged particle \( q \) is \( F = qE=q\frac{\sigma}{\epsilon_0} \).

Step2: Electric Field After Removing One Plate

When one plate is removed, the electric field due to a single charged plate at a point near the plate (where the particle is) is \( E'=\frac{\sigma}{2\epsilon_0} \). But wait, initially, the electric field between the two plates of the capacitor is the superposition of the electric fields from both plates. Each plate contributes \( \frac{\sigma}{2\epsilon_0} \) to the field between the plates, so the net field \( E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0} \).

After removing one plate, the electric field at the location of the particle (which was between the two plates) will now be due to the remaining single plate. But the charge on the remaining plate is still \( Q \) (assuming the capacitor had charge \( Q \) on each plate initially), and the surface charge density \( \sigma=\frac{Q}{A} \) (where \( A \) is the area of the plate).

The force on the charged particle initially is \( F = qE=q\frac{\sigma}{\epsilon_0} \). After removing one plate, the electric field at the position of the particle (which is now in the field of a single charged plate) is \( E'=\frac{\sigma}{2\epsilon_0} \)? Wait, no. Wait, when the two plates are present, the field between them is \( E=\frac{\sigma}{\epsilon_0} \) (because the fields from both plates add up). When we remove one plate, the field at the location of the particle (which was between the two plates) will be due to the remaining plate. But the key point is that the charge on the remaining plate is still \( \sigma A \), and the electric field at a point near the plate (on the side where the other plate was) is \( \frac{\sigma}{2\epsilon_0} \)? No, that's a mistake. Wait, actually, when the two plates are charged with \( +\sigma \) and \( -\sigma \), the field between them is \( E = \frac{\sigma}{\epsilon_0} \), and outside the plates, the field is zero. When we remove the negative plate (for example), the remaining positive plate will have a field of \( \frac{\sigma}{2\epsilon_0} \) on both sides (inside and outside the region where the other plate was). But the charged particle was in the region between the two plates. Initially, the force was \( F = qE=q\frac{\sigma}{\epsilon_0} \). After removing one plate, the electric field at the particle's position is now \( \frac{\sigma}{2\epsilon_0} \)? Wait, no, that's incorrect. Let's think again.

Wait, the electric field due to a single infinite charged plate is \( E=\frac{\sigma}{2\epsilon_0} \) on both sides of the plate. When we have two plates, one with \( +\sigma \) and one with \( -\sigma \), the field between them is \( \frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0} \) (because the fields from both plates are in the same direction between the plates), and outside the plates, the fields cancel out (\( \frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0} = 0 \)).

Now, when we remove one plate (say the negative plate), the remaining positive plate will produce a field of \( \frac{\sigma}{2\epsilon_0} \) at the position of the particle (which was between the two plates earlier). But initially, the force was \( F = q\times\frac{\sigma}{\epsilon_0} \). After removing the plate, the force is \( F'=q\times\frac{\sigma}{2\epsilon_0}=\frac{F}{2} \)? Wait, no, that can't be right. Wait, maybe I made a mistake in the charge distribution.

Wait, the charge on the capacitor plates: when a capacitor is charged, the charge on each plate is \( Q \), so \( \sigma=\frac{Q}{A} \). When we remove one plate, the charge on the remaining plate is still \( Q \), so \( \sigma=\frac{Q}{A} \). The electric field at a point near the remaining plate (on the side where the other plate was) is \( E'=\frac{\sigma}{2\epsilon_0} \). But initially, the electric field between the two plates was \( E = \frac{\sigma}{\epsilon_0} \) (because the two plates have fields that add up). So the force initially is \( F = qE=q\frac{\sigma}{\epsilon_0} \), and after removing one plate, the force is \( F'=qE'=q\frac{\sigma}{2\epsilon_0}=\frac{F}{2} \)? Wait, but that's option 2. But wait, another way: the electric field between the plates of a capacitor is \( E=\frac{V}{d} \), and \( C = \frac{\epsilon_0 A}{d} \), \( Q = CV \). When we remove one plate, the system is no longer a capacitor, but a single charged plate. The electric field at a distance \( x \) from the plate (where \( x\ll\sqrt{A} \), so we can consider it as an infinite plate) is \( E=\frac{\sigma}{2\epsilon_0} \). Initially, between the two plates, \( E=\frac{\sigma}{\epsilon_0} \) (since the two plates' fields add). So the force \( F = qE = q\frac{\sigma}{\epsilon_0} \), and after removal, \( F'=q\frac{\sigma}{2\epsilon_0}=\frac{F}{2} \).

Answer:

1. \( \frac{F}{2} \)