# How will the capacitance of a capacitor reduce when the distance between the plates is increased?

Now I know that if the potential difference between the plates increase that capacitance will reduce, but for that also I thought as the distance between the plates increases, the electric field strength between them reduces and since the field strength is reduced that would mean that the potential difference between the plates is reduces. But it is seen that as distance increases, the potential difference increases (which I don't understand how, so please help me in that also). I know that we can prove capacitance is indirectly proportional to distance by the formula $$C=\epsilon A/d$$, but I am not understanding how the relationship between $$C$$ and $$d$$ came in place. Practically, it can be proven by a circuit, but I need a theoretical explanation.

Now I know that if the potential difference between the plates increase that capacitance will reduce

Ideally, this is not the case. Capacitance depends on the geometry of the conductors, not the applied voltage.

It's a common confusion that $$C\equiv Q/V$$ shows an inverse relation between $$C$$ and $$V$$; however, you're forgetting that in order to change the potential difference $$V$$ you must also change the charge $$Q$$ on each plate. $$Q$$ and $$V$$ will change together in such a way that their ratio is constant.

also I thought as the distance between the plates increases, the electric field strength between them reduces and since the field strength is reduced that would mean that the potential difference between the plates is reduces.

In the idealization of infinite parallel plates, if we hold the charge constant as the plates separate, the field will not change. Instead, due to the increased distance, the potential difference will increase and the capacitance will decrease.

If instead the potential is held constant, then as the plates separate the charge separation will decrease, again lowering the capacitance. We get the same conclusion either way because all that matters is the geometry.

But it is seen that as distance increases, the potential difference increases (which I don't understand how, so please help me in that also).

Yes, if the charge is held constant, we approximate the field as uniform, and an increased distance means a larger potential difference. It's analogous to how if you raise a ball higher and higher into the air, you are increasing the potential energy of the ball-Earth system.

I know that we can prove capacitance is indirectly proportional to distance by the formula $$C=\epsilon A/d$$, but I am not understanding how the relationship between $$C$$ and $$d$$ came in place.

It should be clear now based on what is described above. More formally, we know the field between the plates is $$E=\sigma/\epsilon=Q/(\epsilon A)$$, so then the potential difference between the plates is $$V=Ed$$, giving a capacitance of $$C=Q/V=Q/(Ed)=\epsilon A/d$$.

Consider a capacitor with capacitance $$C$$ which has a charge $$Q$$ when the potential difference is $$V$$ and $$Q=CV$$.

Now assume a second identical capacitor with charge $$Q$$ and potential difference $$V$$.

Now place the positive plate of one capacitor onto the negative plate of the other capacitor which means that the net charge on those two plates is zero.

Those two connected plates can now be removed with no change in the distribution of charges and voltages.

What have you got left?
A capacitor with a separation of twice that of one of the original capacitors, with a potential difference of $$2V$$ between the plates and charge stored $$Q$$, ie a capacitor with a capacitance which is half that of one of the original capacitors.

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Another way of looking at the situation is to note that the electric field between the two plates is $$\frac{Q}{A\epsilon}$$ and also $$\frac Vd$$.
If the charge stored is kept the same then the electric field must stay the same if the separation is doubled, so the potential difference must double, ie the capacitance has halved.

Consider a charged, insulated capacitor. One plate carries Q1=Q and the other Q2=-Q. If you increase the distance between the plates you are increasing the distance between Q1 and Q1. This will increase the potential energy P. In the case of charged plates the energy increases linearly with distance if they are not too far apart. Thus V=P/Q increases with d and C=Q/V decreases with 1/d.

Physically, the Capacitance of the plates at a position is the magnitude of charge given to the plates to maintain a potential difference of 1 Volt. If the distance between the plates increases, the potential difference increases because the magnitude of the electric field between them is roughly the same. To, maintain a potential difference of 1 Volt, less magnitude of charge should be given which implies a lesser value of capacitance.

"I know that if the potential difference between the plates increase that capacitance will reduce" This is not generally true...

Capacitance is defined as $$C=\tfrac QV$$ in which $$Q$$ is the magnitude of charge (equal and opposite) on either plate and $$V$$ is the pd between the plates.

For a capacitor of fixed plate area, $$A$$, fixed plate separation and a linear dielectric, we can show, using Gauss's law, that $$C$$ is a constant, independent of $$V$$.

Suppose, though, that we place equal and opposite charges charges, $$±Q$$ on the plates, disconnect the plates from anything else so that they retain their charges and then increase the plate separation. For a parallel plate capacitor Gauss's law shows that (provided the separation remains much smaller than the plate dimensions), at each point in the gap between the plates the electric field strength stays the same (magnitude $$E=\frac Q{\epsilon_0 A}$$). Therefore $$V$$ increases as more work needs to be done per unit charge transferring a test charge from one plate to the other over the greater distance. So from the definition, $$C=\frac QV$$, the capacitance decreases, in accordance with the parallel plate capacitance formula, $$C=\tfrac {\kappa\epsilon_0 A}d$$.

Suppose, though, that, instead of holding $$Q$$ constant, we hold $$V$$ constant, for example by leaving a battery connected across the plates, while we slowly increase their separation. The electric field strength at all points in the gap will decrease and Gauss's law tells us that the magnitude of charge on the plates will decrease (as $$E=\frac Q{\epsilon_0 A}$$). This happens by means of charge flowing 'backwards' through the battery. So we reach the same conclusion: the capacitance, defined by $$C=\frac QV$$, decreases.

I know that we can prove capacitance is indirectly proportional to distance by the formula $$C=\epsilon A/d$$, but I am not understanding how the relationship between $$C$$ and $$d$$ came in place. Practically, it can be proven by a circuit, but I need a theoretical explanation.

The following ia an attempt at a more intuitive explanation.

Start with the term "capacitance" of a device. It means the capability of the device to hold net charge on its plates for a given voltage across the plates. The greater the charge per volt, the greater the ability (capacitance) of the device for holding charge. Thus the following relationship:

$$C=\frac{Q}{V}\tag{1}$$

Since the plates have equal and opposite charge, we know there is a force of attraction between the plates. If the plates were separated by air or a vacuum and a positive test charge $$Q$$ were placed between the plates, that charge would experience a force causing it to accelerate towards the negatively charged plate. By definition, the magnitude of that force would be

$$F=EQ\tag{2}$$

Where $$E$$ is the magnitude of the electric field between the plates.

Now consider what happens if the plates are brought closer together while the voltage is maintained. The force of attraction between the plates increases, and thus the force on the test charge between the plates increases. This, in turn, tells us that the electric field has become stronger, yeidling the following relationship for a parallel plate capacitor:

$$E=\frac{V}{d}\tag{3}$$

Next consider the relationship between the electric field, the surface charge density $$\sigma$$ on the plates, and the permittivity $$\epsilon$$ of the medium on the plates:

$$E=\frac{\sigma}{\epsilon}=\frac{Q}{A\epsilon}\tag{4}$$

Intuitively, the greater the surface density of the charge on the plates the stronger the electric field between them, everything else being equal.

Substituting the rights side of eq (4) for $$E$$ in eq (3) gives:

$$V=\frac{Qd}{A\epsilon}\tag{5}$$

Finally, substituting $$V$$ from eq (5) for $$V$$ in eq (1) yields

$$C=\frac{A\epsilon}{d}\tag{6}$$

Hope this helps.