Now, if I want to charge the capacitor, this means pumping charges on
one of the plates which, by induction, produces an equal but opposite
charge on the opposite plate.
The charge on the opposite plate is not produced by induction. An external voltage source, such as a battery, does work to remove electrons from one plate and deposit them on the other, resulting in a net negative charge on one plate and net positive charge on the other. The total electrical charge (protons and electrons) on the plates remain unchanged. The charges are just redistributed establishing a uniform electric field between the plates and a potential difference equal to the product of the electric field and separation of the plates.
Electrical potential energy is supposedly stored because it takes work
to move charge against the electric field (and in fact equal to the
work if we set 0 potential energy to an uncharged state).
Essentially correct.
As an analogy, I imagine a capacitor as a spring where “compressing
the spring” means adding charge.
The ideal spring is the mechanical analogue of a capacitor, as discussed further below.
But in the scenario I have presented, there is NO electric field
outside the space between the plates. There is no repulsive force. No
“spring force” can kick out charges.
Neglecting fringe effects, there is no electric field in the space outside the capacitor due to the field between the plates. But the charges are moved from one plate to the other by an externally applied field of an external voltage source, such as a battery.
The positive terminal of the battery pulls electrons off the plate connected to it. The negative terminal of the battery deposits and equal number of electrons on the plate attached to it. It requires a force to perform work to pull and push charges from and to the plates. That force is supplied by electric field of the battery.
How can we store electric potential energy if there is no force (due
to above reason) repelling our efforts to accumulate charges on a
conducting plate? Is there anything wrong with my reasoning?
What's wrong is you are not considering the external electrical force supplied by the battery to move charges against the attraction or repulsive forces. In this regard, perhaps the ideal spring analogy will help.
SPRING ANALOGY:
An ideal spring is the mechanical analogue of an ideal capacitor (and vice versa). The analogous parameters are:
- Voltage $V$ applied to the capacitor is analogous to the force $F$ applied to the spring.
- The displacement $x$ of the spring is analogous to the charge $Q$ delivered to the capacitor.
- The capacitance $C$ of the capacitor is analogous to the inverse of the spring constant $k$ of the spring.
I need to emphasize that charge does not equal displacement, voltage does not equal force, and capacitance does not equal the inverse of the spring constant. These are analogies, not identities.
The electrical potential energy stored in the electric field of the charged capacitor is commonly shown as
$$E_{C}=\frac{CV^2}{2}$$
The relationship between voltage, capacitance, and charge for a capacitor is
$$V=\frac{Q}{C}$$
Substituting this in the previous equation we obtain
$$E_{C}=\frac{Q^2}{2C}$$
The elastic potential energy stored in a spring that is compressed (or extended) a displacement of $x$ is given by
$$E_{S}=\frac{kx^2}{2}$$
Demonstrating that $C$ is analogous to the inverse of $k$ and $Q$ is analogous to $x$.
Since for a spring, $F=kx$, or $x=F/k$, we can express the elastic potential energy stored in the spring as
$$E_{S}=\frac{F^2}{2k}$$
Comparing that to the first equation for the energy stored in a capacitor shows that the spring force $F$ is analogous to the capacitor voltage, $V$.
If two springs with different spring constants are subjected to identical forces, more elastic potential energy will be stored in the spring with the lower spring constant. This is because the stored elastic potential energy varies as the square of the displacement and linearly with the spring constant. So if the spring constant of spring 1 is half the spring constant of spring 2, for the same force the displacement $x$ of spring 1 will be twice that of spring 2. The elastic potential energy stored in spring 1 will be twice that of spring 2.
One final point. As charge is moved from one plate to the other it becomes more and more difficult to move additional charge. That's because when the initial charge is moved the plates are essentially neutral so that little work is required. This explains why during the initial phase of charging a capacitor the current (rate of charge delivery) is maximum.
However as net charge builds up, the attraction and repulsion forces increase resisting the transfer of additional charge. So now the current (rate of charge delivery) is decreasing as the voltage across the capacitor builds.
Once the voltage across the capacitor equals the source (e.g. battery) voltage, the net driving voltage becomes zero and current cease. The capacitor is fully charged.
The spring analogy is the force required to cause displacement increases as the amount of displacement increases.
Hope this helps.
