What happens when an uncharged object is brought into an electric field?

[aspodkfpo]

Homework Statement

What happens when an uncharged object is brought into an electric field? Textbook states that a charge is induced within the object and so it experiences the force. Isn't the object still neutral, why would it experience a force?

*Attempt at solution box isn't working.

Relevant Equations

none

Tecd

 

[haruspex]

Homework Statement:: What happens when an uncharged object is brought into an electric field? Textbook states that a charge is induced within the object and so it experiences the force. Isn't the object still neutral, why would it experience a force?

*Attempt at solution box isn't working.
Relevant Equations:: none

Tecd

If it is a uniform field then you are right, there is no net force. But say it is from an isolated positive charge. Within the body, negative charges will move towards it and positive charges away. The negative charges are now closer to the isolated charge than the repulsed positive charges are, so experience the greater force.

 

[rude man]

Another example: take a parallel plate capacitor, put Q on one plate and -Q on the other (no externally applied emf).

Slowly move a third, uncharged plate of finite thickness and area equal to the capacitor plates into the gap. There is a force acting to pull the third plate into alignment with the two capacitor plates.

You can compute quantitatively (ignoring E field fringing effects, or plate area >> plate separation) by invoking virtual work. The E field is unchanged after the third plate is inserted but the volume per unit area is reduced due to the thickness of the third plate, so total field energy after the insertion is less than total field energy before insertion.

(BTW the same thing happens when you move a dielectric into the capacitor).

 

[aspodkfpo]

Another example: take a parallel plate capacitor, put Q on one plate and -Q on the other (no externally applied emf).

Slowly move a third, uncharged plate of finite thickness and area equal to the capacitor plates into the gap. There is a force acting to pull the third plate into alignment with the two capacitor plates.

You can compute quantitatively (ignoring E field fringing effects, or plate area >> plate separation) by invoking virtual work. The E field is unchanged after the third plate is inserted but the volume per unit area is reduced due to the thickness of the third plate, so total field energy after the insertion is less than total field energy before insertion.

(BTW the same thing happens when you move a dielectric into the capacitor).

? How am I thinking about this wrong? E = V/d, d decreases and E becomes larger.

 

[rude man]

? How am I thinking about this wrong? E = V/d, d decreases and E becomes larger.

There is no more "d" after the central plate is inserted. There are two smaller "d"s, each smaller than d/2 since the central plate has finite thickness ##\delta##.

The fact that the E field does not change can be ascertained by showing that the charge densities on each inside face = Q/A or -Q/A, A = plate area, same as before, so that ## Q/(\epsilon_0 A) = ## E, the same before & after insertion.

Remember that we are ignoring fringing effects so if you know ##\sigma## on one inside surface you know that the E field is ##\sigma/\epsilon_0## all the way from one inside surface to the opposing inside surface. Plate separation enters the problem only if you want to compute potentials, which we don't need to (but just FYI the new V changes from Ed to E(d-##\delta##). )

So anyway you end up with the same E field but with smaller volume, changing from dA to ##(d-\delta)## A.

EDIT: Or you can go as follows:
Let unprimed = before 3rd plate insertion
primed = after insertion
Then,
## Q'=Q \Rightarrow C'V' = CV. ##
## C = \epsilon_0 A/d ##
## V = Ed ##
## C' = \epsilon_0 A/(d-\delta) ## (2 capacitors in series)
## V' = 2E'(d-\delta)/2 = E'(d-\delta) ##
So ## C'V' = CV \Rightarrow E'=E. ##