The area of the plates of a parallel plate ca | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$E=\frac{\sigma}{K \epsilon_{o}}$

$\frac{d v}{d y}=\frac{\sigma}{K \epsilon_{o}}$

$\int_{0}^{v} d v=\frac{\sigma}{\lambda \epsilon_{o}} \int_{0}

Vedclass · Accepted Answer

$\pi \varepsilon_0\lambda A / 2d$

Vedclass · Answer

$E=\frac{\sigma}{K \epsilon_{o}}$

$\frac{d v}{d y}=\frac{\sigma}{K \epsilon_{o}}$

$\int_{0}^{v} d v=\frac{\sigma}{\lambda \epsilon_{o}} \int_{0}^{d} \cos \left(\frac{\pi y}{2 d}ight) d y V$

$=\frac{\sigma}{\lambda \epsilon_{o}} 	imes \frac{2 d}{\pi}\left[\sin \frac{\pi y}{2 d}ight]_{0}^{d}$

$=\frac{\sigma}{\lambda \epsilon_{o}} 	imes \frac{2 d}{\pi}$

$C=\frac{Q}{V}=\frac{A \lambda \epsilon_{o} \pi}{2 d}$

Similar Questions

The dielectric constant $k$ of an insulator cannot be

With the rise in temperature, the dielectric constant $K$ of a liquid

What is polarisation ?

The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out