# force variation by plate of parallel plate capacitor

Now If I consider a parallel plate capacitor connected across a battery then ,

force(f) between the plates should not be proportional to distance(d) (that's what I think)

as force due to plate on other=$q\sigma/\epsilon_0$ where $\sigma$would be the charge density and q be the charge on other plate.

But my text says it's inversely proportional to distance . Now what I think is that what I did above may be correct for only an isolated capacitor but given is connected across terminals; If so then how can I derive it mathematically that : $f\alpha d^-2$

## 1 Answer

Remember that the capacitance is proportional to $\frac{1}{d}$. This means that, when the voltage is constant, the charge (and the charge density) will both be proportional to $\frac{1}{d}$.

Can you see it now?

• I see that charge would be proportional $d^-1$ clearly but how come force ? – Physicsapproval Jan 18 '17 at 2:47
• Surface charge determines the electric field, Force is the product of charge and electric field - so you should find the force is inversely proportional to the square of the distance. Details of the derivation can be found here – Floris Jan 18 '17 at 3:05