A small object is released at point 3in the center of the capacitor in FIGURE Q23.11. For each situation, does the object move to the right, to the left, or remain in place? If it moves, does it accelerate or move at constant speed?

a. A positive object is released from rest.

b. A neutral but polarizable object is released from rest.

c. A negative object is released from rest.

Short Answer

Expert verified

(a) A positive object at the center will move to the negative plate and accelerate towards it.

(b) A neutral but polarizable object will remain at rest.

(c) A negative object at the center will move to the positive plate and accelerate towards it.

Step by step solution

01

Given information (part a)

Given the figure of the plate of a parallel plate capacitor

02

Explanation (part a)

The electric field strength of a parallel plate capacitor is given as

E=σεoq

The direction of the parallel plate capacitor is in the direction of the positive plate to the negative plate.

When a positive charge is released from rest from the center of the plate it will move in the direction of the electric field line and accelerates in the direction field line. Therefore the positive charge will move towards the negative plate and accelerates.

03

Given information (part b)

Given the figure of the plate of a parallel plate capacitor

04

Explanation (part b)

The electric field strength of a parallel plate capacitor is given as

E=σεoq

The direction of the parallel plate capacitor is in the direction of the positive plate to the negative plate.

When a neutral polarizable object moves from the rest it will move until it reaches the equilibrium position where the electric field strength is even, since the neutral object is released from the center of the plate which is an equilibrium state therefore it will not move and remain in the equilibrium position.

05

Given information (part c)

Given the figure of the plate of a parallel plate capacitor

06

Explanation (part c)

The electric field strength of a parallel plate capacitor is given as

E=σεoq

The direction of the parallel plate capacitor is in the direction of the positive plate to the negative plate.

When a negative charge is released from rest from the center of the plate it will move in the direction opposite to the electric field line and accelerates in the opposite direction of the electric field line. Therefore the negative charge will move towards the positive plate and accelerates.

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