# Influence of an external electric field on an uncharged capacitor

Influence of an external electric field on an uncharged capacitor.

Fig A shows an uncharged capacitor with no external electrical field. There is no net charge on each of the plates. Since the net charge on each of the plates is zero, no electric field is produced outside the plates. The dielectric between the plates consists of polar molecules, randomly oriented since there is no electric field in the dielectric.

Fig B shows the same capacitor but now immersed in a uniform electric field. By convention, the direction of the field is the direction of the force that a positive charge would experience if placed in the field. Due to the influence of the field, I believe the charges on the capacitor plates will be redistributed, so that there is now a voltage between the outer surfaces of the plates (between points 1 and 2, separated by distance d) equal to the voltage drop of the field between the points. However, there is still no net charge on either plate. If I draw a closed Gaussian surface around each plate, there will be no field outside the plates due to the charge on the plates. Consequently, it appears that there is no field in the dielectric due to the charges on the plates.

In other words, is my depiction of the situation in Fig B correct with respect to the influence of the external electric field on the dielectric between the plates?

I have added Fig C, which shows the same capacitor as Fig A and B except that it has been charged by a battery and has voltage $$V=Ed$$.

There will be a field E inside the plates. Draw your Gaussian surface to surround only the right side of the left plate, not the whole plate. This gives E between the plates. If you draw it around the whole plate, the surface integral is zero because E enters at the left and leaves out the right side of the plate.

• If there is a field in the dielectric, does that infer there is energy stored in the capacitor? I would think not, since there is no net charge on either plate. Also, does not the field produced in the left plate oppose the external field from the left? – Bob D Feb 4 at 22:01
• If we can ignore the end effects (infinite plates) can't we consider the capacitor plates to be a Faraday Gage? The field induced in the plates by the external field cancels external field so that the field in the dielectric is zero. Maybe I'm wrong. But thanks for your help. – Bob D Feb 4 at 23:05
• 1. The energy inside the capacitor is produced by the external E field which causes the E field inside the capacitor. – Clem Feb 5 at 11:57
• 2. There is no total E field inside either plate, since the field of the surface charges cancels the external field inside (but not outside) the plate. – Clem Feb 5 at 12:01
• 3. It is not a Faraday cage, because the plates are not connected. If the plates were connected, shorting the capacitor, it would no longer be a capacitor. The plates would each become charged, cancelling the external field between the plates. – Clem Feb 5 at 12:02

I believe the charges on the capacitor plates will be redistributed so that there is now a voltage between the outer surfaces of the plates (between points 1 and 2, separated by a distance $$d$$) equal to the voltage drop $$V_{1-2} = Ed$$ of the field between the points.

• Yes, the charges on the capacitor plates will redistribute as illustrated in Fig B under the influence of the $$\vec E_{ext}$$ field.
• The voltage, $$V_{1-2} = Ed =0$$, across the capacitor while it is immersed in the external field. The external field induces charge separation and dipole alignment, which creates an opposing field, resulting in a net field of zero magnitude field at every point inside the capacitor.
1) The field induced in the capacitor plates by charge separation is of equal magnitude to $$\vec E_{ext}$$ and in the opposite direction.
2) The field induced in the dielectric (by dipole alignment and charge separation) is likewise of equal magnitude and opposite direction to the $$\vec E_{ext}$$ field at every point in the dielectric. Thus, the net field within the dielectric is zero at every point.
3) Since capacitor plates (left & right) and dielectric are in series, and the voltage across each of these elements is zero, the voltage measured by the voltmeter across the entire capacitor is zero.

If I draw a closed Gaussian surface around each plate, there will be no field outside the plates due to the charge on the plates. Consequently, it appears that there is no field in the dielectric due to the charges on the plates.

• Yes, the Gaussian surface surrounding each plate contains zero net charge. And yes, the resultant to $$\vec E$$ field external to the capacitor plates is zero.
• But, there is a field in the dielectric. That polarizing field is not due to the capacitor plate charges, rather it is the $$\vec E_{ext}$$ field which penetrates the entire capacitor. Thus, the $$\vec E_{ext}$$ field aligns the dipoles of the dielectric.

In other words, is my depiction of the situation in Fig B correct with respect to the influence of the external electric field on the dielectric between the plates?

• No, the dipoles of the dielectric will align, and the charges of each dipole will be separated by the force of the external field.
• The alignment and separation of the dielectric dipoles produce an electric field which exactly opposes the $$\vec E_{ext}$$ field. The result is a net-zero magnitude $$\vec E$$ field at every point within the dielectric.

Capacitor charged by a Voltage Source:

• Connect a capacitor to a voltage source and current will flow. Electrons accumulate on one plate, and positive charges accumulate on the other. The charges on each plate create an $$\vec E$$ field, which constructively superimposes. This field polarizes the dielectric.

Summary: Capacitor charged by External Field

• In an open circuit capacitor immersed in an external field, there is no movement of charge off of one plate and onto the other.
• The positive and negative charges, within each capacitor plate, are driven to opposite sides of their respective plates. Thus, when the capacitor is immersed in the $$\vec E_{ext}$$ field, the capacitor plates are charged/polarized.
• The dielectric dipoles are aligned and separated because of the force exerted by the $$\vec E_{ext}$$ field.
• At full charge, the net $$\vec E$$ field = $$0$$ at every point inside the capacitor plates and dielectric. Therefore, the voltage across the entire capacitor is zero.