Electrostatics Concept Page - 6

Formula

Electric field produced by a point charge at any point

E⃗ =Q4πϵ0r2r^$\overrightarrow{E}={\displaystyle \frac{Q}{4\pi {ϵ}_0{r}^2}}\hat{r}$
where r^$\hat{r}$ is a unit vector from the origin to the point r.

Formula

Electric field due to an infinite linear charge distribution

We have the charge qenc$q_{enc}$ enclosed by the
cylinder. Because the linear charge density (charge per unit length, remember) is
uniform, the enclosed charge is λ$\lambda$h.Thus, Gauss law
ϵoΦ=qenclosed${ϵ}_o\mathrm{\Phi }=q_{enclosed}$
ϵoE(2πrh)=λh${ϵ}_{o}E(2\pi rh)=\lambda h$
E=λ2πϵ0r$E={\displaystyle \frac{\lambda }{2\pi {ϵ}_{0}r}}$
where λ$\lambda$ is the linear charge density and r is the distance at which the electric field is to be calculated.

Formula

Electric field due to uniformly charged infinite plane sheet

By gauss law
 0

E

:

dA:

qenc,


ϵo(EA+EA)=σA${ϵ}_o(EA+EA)=\sigma A$
E=σ2ϵ0$E={\displaystyle \frac{\sigma }{2{ϵ}_0}}$
where σ$\sigma$ is the surface charge density.

Formula

Electric field due to a charged ring along the axis

E=kQx(x2+R2)32$E={\displaystyle \frac{kQx}{{(x^2+R^2)}^{\frac{3}{2}}}}$
where Q=2πλR$Q=2\pi \lambda R$
R$R$ is the radius of the ring
λ$\lambda$ is the charge density
x$x$ is the distance from the centre of the ring along the axis

Formula

Electric field due to a charged semicircle

Ex=0$E_x=0$
Ey=λ2πϵ0a$E_y={\displaystyle \frac{\lambda }{2\pi {ϵ}_{0}a}}$
E=E2x+E2y−−−−−−−√=Q2π2a2ϵ0$E=\sqrt{E_x^2+E_y^2}={\displaystyle \frac{Q}{2{\pi }^2{a}^2{ϵ}_0}}$
Where Q=λ(πa)$Q=\lambda (\pi a)$

Formula

Electric field due to a uniformly charged disc

E=kσ2π[1−zz2+R2−−−−−−√]$E=k\sigma 2\pi [1-{\displaystyle \frac{z}{\sqrt{z^2+R^2}}}]$
where k=14πϵ0$k={\displaystyle \frac{1}{4\pi {ϵ}_0}}$ and σ$\sigma$ is the surface charge density

Formula

Electric field due to a uniformly charged thin spherical shell

Figure shows a charged spherical shell of total charge q and radius R and two
concentric spherical Gaussian surfaces, S1 and S2.  as we applied Gauss law to surface S2,for which r≥R$r\ge R$,we would find that
E=Q4πϵ0r2$E={\displaystyle \frac{Q}{4\pi {ϵ}_0{r}^2}}$                       
 Applying Gauss law to surface S1, for which r<R, leads directly to
E=0$E=0$                                    (r<R)$(r<R)$
where r is the distance from the centre to outside of the shell.

Formula

Electric field due to a conducting cylinder

For r≥R$r\ge R$
E=λ2πϵ0r$E={\displaystyle \frac{\lambda }{2\pi {ϵ}_{0}r}}$
For r<R$r<R$
E=0

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