# Why decreasing the distance between a parallel plate capacitor increases the electric field? Wouldn't it remain the same? $\sigma/\epsilon_0$? [closed]

I have been learning capacitors and came across the formula for the capacitance of a parallel plate capacitor. In it capacitance is inversely proportional to the distance between the plates because decreasing distance increases electric field is what I have read. But by the formula Sigma/epsilon naught which is independent of distance. I need help in clearing this. I am a high school student so pls keep it simple.

• You need to specify what the capacitor is connected to, if anything.
– Puk
Commented Dec 5, 2023 at 6:23

Two possibilities.

1 Charged capacitor not connected to anything else.
Charge,$$q$$, and hence charge density, $$\sigma = q/A$$, cannot change.
Electric field $$E = \sigma/\epsilon_0 = V/d$$ does not change.
Work needs to be done changing the separation of the plates which in turn proportionately changes the potential difference across the plates, $$V\propto d$$.

2 Charged capacitor connected to battery.
As the potential difference across the plates is constant work done, $$\int \vec E\cdot d\vec \ell$$, in moving unit charge between plates is constant.
If the separation of the plates is decreased then the electric field must increase to keep $$\int \vec E\cdot d\vec \ell$$ constant.
As $$E = \sigma/\epsilon_0$$, if $$E$$ increases so must the surface charge density, $$\sigma$$, and hence the charge on the capacitor plates.

• precise and concise,+1 Commented Dec 5, 2023 at 17:11

Assuming a constant potential difference is applied to capacitor, Like battery

Why do you think that The Charge density "sigma" will not change? If you calculate carefully, The electric field for parallel plate capacitor is $$\frac{Q}{Ak_0}$$

As distance is decreased, Charge Increases on each plate(but with opposite signs),Which in turn increases electric field

Why does charge increase, When distance is decreased?

The potential difference $$E\cdot d$$ equals $$\frac{Qd}{Ak_0}$$,Now if this is maintained constant by the battery, And distance is decreased, Charge must increase, As they are inversely proportional

So concluding

When distance is decreased, to maintain constant potential difference, Magnitude of charge on each plate is increased by the battery, Which leads to increase in electric field

• I get it but how do we explain with the formula Sigma/epsilon naught? It stays the same, whatever the distance is? Commented Dec 5, 2023 at 16:17
• @Dhyaneshwar,What is Sigma?Charge density,isn't it? so charge changes,Charge density changes,Sigma changes Commented Dec 5, 2023 at 16:44

Starting from the expression of the electric field of a 2-dimensional circular charged plate with radius R. The correct expression for the electric field is $$\mathbf{E}(x)=\frac{|\sigma|}{\epsilon_0\epsilon_r}(1-\frac{x}{\sqrt{x^2+R^2}})\hat{x}$$ where $$x$$ indicates the distance frome the plate. If the plate is infinite $$R\rightarrow \infty$$ or the distance $$x\rightarrow 0$$ reduces to $$\mathbf{E_{max}}=\frac{|\sigma|}{\epsilon_0\epsilon_r}\hat{x}=(constant)\hat{x}$$ So, considering a finite plate and a finite distance $$x$$ it increases as $$x$$ approaches zero, where the electric field reach it's maximum value. The field would be uniform only if the plate is infinite.

Thus, the capacity is x-dependent and increas as x decreases $$\Delta V=\int_0^x\mathbf{E(x')}\cdot dx'$$ $$C(x)=\frac{Q}{\Delta V}=\frac{Q}{\int_0^x\mathbf{E(x')}\cdot dx'}=\frac{Q}{\frac{|\sigma|}{\epsilon_0\epsilon_r}\int_0^x(1-\frac{x'}{\sqrt{x'^2+R^2}})\cdot dx'}=\frac{Q\epsilon_0\epsilon_r}{|\sigma|}\frac{1}{x-\sqrt{x^2+R^2}-R}$$

The capacitance has nothing to do with electric field, it is constant for a given capacitor.