Ch.2 Electric Field

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Electric Field Models

Electric Field of Point Charges

E⃗=14πϵ0qr2r⃗\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\vec{r}
ϵ0=8.85⋅10−12 C2/N m2\epsilon_0=8.85\cdot10^{-12}\text{ C}^2/\text{N m}^2 is the permittivity constant

With multiple charges, F⃗on q=∑iF⃗i on q\vec{F}_{\text{on }q}=\sum_i\vec{F}_{i\text{ on }q}
Similarly, electric field is the vector sum of individual fields.

Consider a dipole with small distance ss between charges of +q+q and −q-q. It has an electric field. At a point $y\gg
s$, an approximation can be made to find the far field of the dipole
(Edipole)y≈14πϵ02qsy3(E_{\text{dipole}})_y\approx\frac{1}{4\pi\epsilon_0}\frac{2qs}{y^3}
In particular, the dipole moment
p⃗=qs, from negative to positive charge\vec{p}=qs,\text{ from negative to positive charge}
is useful. The magnitude p=qsp=qs determines the field strength, and has SI units C m\text{C m}. Using this, the far field on the axis of the dipole moment
E⃗dipole=14πϵ02p⃗r3\vec{E}_{\text{dipole}}=\frac{1}{4\pi\epsilon_0}\frac{2\vec{p}}{r^3}
where rr is the distance from the center of the dipole

Electric Field Lines

instead of vectors, use electric field lines

Electric Field of Continuous Charge Distributions

Let QQ be the total charge (lowercase qq is for individual point charges) of an object with length LL. The line charge densityλ\lambda is defined as
λ=QL\lambda=\frac{Q}{L}
with units C/m\text{C}/\text{m}
For a charged surface of area AA, we define surface charge densityη\eta to be
η=QA\eta=\frac{Q}{A}
with units C/m2\text{C}/\text{m}^2. (Volume covered in Ch.3)
We assume all objects are uniformely charged unless otherwise noted.

Using integration, the total charge becomes
Q=∫objectdQQ=\int_\text{object}dQ
To integrate over a rod of length LL, set the origin of the xx-axis at one end and divide into small segments of size dxdx. Then due to the linear charge density, we have that dQ=λdxdQ=\lambda dx. We integrate from x=0x=0 to x=Lx=L, the length of the rod, so the integral becomes
Q=∫0LλdxQ=\int_0^L\lambda dx
If λ\lambda is constant, it can be factored out to yield Q=λLQ=\lambda L. But if λ\lambda were a function of xx, integration can be used to find LL.

Electric Field of a Rod


Since this is symmetric, the vertical component of E⃗\vec{E} is 00. It suffices to calculate E⃗x\vec{E}_x
The xx-component of the field of segment ii is
(E⃗i)x=E⃗icos⁡θi=14πϵ0ΔQri2cos⁡θi(\vec{E}_i)_x=\vec{E}_i\cos\theta_i=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q}{r_i^2}\cos\theta_i
From the figure, we find that ri2=yi2+x2r_i^2=y_i^2+x^2 and cos⁡θi=x/ri=x/(yi2+x2)1/2\cos\theta_i=x/r_i=x/(y_i^2+x^2)^{1/2}. Thus, the field of segment ii is
(E⃗i)x=14πϵ0ΔQyi2+x2xyi2+x2=14πϵ0xΔQ(yi2+x2)3/2(\vec{E}_i)_x=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q}{y_i^2+x^2}\frac{x}{\sqrt{y_i^2+x^2}}=\frac{1}{4\pi\epsilon_0}\frac{x\Delta Q}{(y_i^2+x^2)^{3/2}}
Assuming the charge density is a constant λ=Q/L\lambda=Q/L, we have ΔQ=λΔy=(Q/L)Δy\Delta Q=\lambda \Delta y=(Q/L)\Delta y. As the size gets smaller, the sum approaches
(E⃗i)x=∫−L/2L/214πϵ0xQdyL(y2+x2)3/2=Qx4πϵ0L∫−L/2L/2dy(y2+x2)3/2(\vec{E}_i)_x=\int_{-L/2}^{L/2}\frac{1}{4\pi\epsilon_0}\frac{xQdy}{L(y^2+x^2)^{3/2}}=\frac{Qx}{4\pi\epsilon_0L}\int_{-L/2}^{L/2}\frac{dy}{(y^2+x^2)^{3/2}}

Evaluation

Set y=xtan⁡(u)⟹dy=xsec⁡2(u)duy=x\tan(u)\implies dy=x\sec^2(u)du
∫xsec⁡2(u)dux3(tan⁡2(u)+1)3/2=1x2∫dusec⁡(u)=sin⁡(u)x2\int\frac{x\sec^2(u)du}{x^3(\tan^2(u)+1)^{3/2}}=\frac{1}{x^2}\int\frac{du}{\sec(u)}=\frac{\sin(u)}{x^2}
From tan⁡(u)=y/x\tan(u)=y/x, we derive sin⁡(u)=y/y2+x2\sin(u)=y/\sqrt{y^2+x^2}
sin⁡(u)x2=yx2y2+x2\frac{\sin(u)}{x^2}=\frac{y}{x^2\sqrt{y^2+x^2}}

(E⃗i)x=Qx4πϵ0L∣yx2y2+x2∣−L/2L/2=14πϵ0⋅QxL2/4+x2(\vec{E}_i)_x=\frac{Qx}{4\pi\epsilon_0L}\left|\frac{y}{x^2\sqrt{y^2+x^2}}\right|_{-L/2}^{L/2}=\frac{1}{4\pi\epsilon_0}\cdot\frac{Q}{x\sqrt{L^2/4+x^2}} So, for any rod with a uniform charge distribution, the field at the point distance rr away that bisects the rod is Erod=14πϵ0∣Q∣rr2+(L/2)2E_\text{rod}=\frac{1}{4\pi\epsilon_0}\frac{|Q|}{r\sqrt{r^2+(L/2)^2}}

If the rod has infinite length, or can be approximated as having infinite length, then we factor out the L/2L/2 in the square root to get
Erod=14πϵ0∣Q∣r⋅L/211+4r2/L2=14πϵ02∣λ∣r1+4r2/L2E_\text{rod}=\frac{1}{4\pi\epsilon_0}\frac{|Q|}{r\cdot L/2}\frac{1}{\sqrt{1+4r^2/L^2}}=\frac{1}{4\pi\epsilon_0}\frac{2|\lambda|}{r\sqrt{1+4r^2/L^2}}
We used the fact that λ=Q/L\lambda=Q/L. Now, letting L→∞L\to\infty we simply have
Erod=14πϵ02∣λ∣rE_\text{rod}=\frac{1}{4\pi\epsilon_0}\frac{2|\lambda|}{r}

Electric Field of a Ring of Charge


The ring of radius RR lies in the xyxy-plane and the point on the zz-axis. Since the component of the field along the xx and yy axes cancel out, we only calculate the zz component, EzE_z. The zz-component of the field due to segment ii is
(Ei)z=Eicos⁡θi=14πϵ0ΔQri2zri(E_i)_z=E_i\cos\theta_i=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q}{r_i^2}\frac{z}{r_i}
Regardless of the segment, rir_i has the same value
ri=R2+z2r_i=\sqrt{R^2+z^2}
Thus,
(Ei)z=14πϵ0z(R2+z2)3/2ΔQ(E_i)_z=\frac{1}{4\pi\epsilon_0}\frac{z}{(R^2+z^2)^{3/2}}\Delta Q
The total field is the sum of the EiE_is
Ez=14πϵ0z(R2+z2)3/2∑i=1NΔQE_z=\frac{1}{4\pi\epsilon_0}\frac{z}{(R^2+z^2)^{3/2}}\sum_{i=1}^N\Delta Q
Everything but ΔQ\Delta Q can be factored out because they are all constants, and ∑ΔQ=Q\sum \Delta Q=Q, the total charge. Thus,
Ez=14πϵ0Qz(R2+z2)3/2E_z=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}

As z≫Rz\gg R

As the distance grows larger, the ring should approach a point charge.
If z≫Rz\gg R
Ez=14πϵ0Qz(R2+z2)3/2≈14πϵ0Qz(z2)3/2=14πϵ0Qz2E_z=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}\approx\frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2)^{3/2}}=\frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}
This is the same as the field of a point charge at distance rr.

The field is 00 at the center of the circle, positive in the positive zz-direction, negative in the opposite.

The field reaches its maximum magnitude around ∣z∣≈R|z|\approx R

Max field strength

Find the value of zz that maximizes the EzE_z equation.
ddz(Ez)=14πϵ0Q(R2+z2)3/2−32QzR2+z2(2z)(R2+z2)3\frac{d}{dz}(E_z)=\frac{1}{4\pi\epsilon_0}\frac{Q(R^2+z^2)^{3/2}-\frac{3}{2}Qz\sqrt{R^2+z^2}(2z)}{(R^2+z^2)^3}
=14πϵ0QR2+z2−3z2(R2+z2)3/2\\=\frac{1}{4\pi\epsilon_0}Q\frac{R^2+z^2-3z^2}{(R^2+z^2)^{3/2}}
Since we may assume R>0R>0, the denominator is never 00. Thus, the critical points are at when
R2−2z2=0⟹z2=R22⟹z=±22RR^2-2z^2=0\implies z^2=\frac{R^2}{2}\implies z=\pm\frac{\sqrt2}{2}R

Electric Field of a Disk of Charge


Above is a thin disk of radius RR with uniformely distributed charge QQ.
The surface charge density is
η=QA=QπR2\eta=\frac{Q}{A}=\frac{Q}{\pi R^2}
Same as above, we calculate the field at a point on the zz-axis.
Since we know the field due to a circle of charge, we divide the disks into small rings of radius rr and width Δr\Delta r.
Each ring ii with radius rir_i contributes to the field
(Ei)z=14πϵ0ΔQiz(z2+ri2)3/2(E_i)_z=\frac{1}{4\pi\epsilon_0}\frac{\Delta Q_iz}{(z^2+r_i^2)^{3/2}}
Thus, the total field is
Ez=∑i=1N(Ei)z=z4πϵ0∑i=1NΔQi(z2+ri2)3/2E_z=\sum_{i=1}^N(E_i)_z=\frac{z}{4\pi\epsilon_0}\sum_{i=1}^N\frac{\Delta Q_i}{(z^2+r_i^2)^{3/2}}
From the surface charge density, the charge ΔQi=ηΔA=2πηriΔr\Delta Q_i=\eta\Delta A=2\pi\eta r_i\Delta r
From this we get
(Edisk)z=ηz2ϵ0∑i=1NriΔr(z2+ri2)3/2(E_\text{disk})_z=\frac{\eta z}{2\epsilon_0}\sum_{i=1}^N\frac{r_i\Delta r}{(z^2+r_i^2)^{3/2}}
As N→∞N\to\infty this approaches the integral
(Edisk)z=ηz2ϵ0∫0Rrdr(z2+r2)3/2(E_\text{disk})_z=\frac{\eta z}{2\epsilon_0}\int_0^R\frac{rdr}{(z^2+r^2)^{3/2}}
Substituting u=z2+r2⟹du=2rdru=z^2+r^2\implies du=2rdr and r:0→R⟹z:z2→z2+R2r:0\to R\implies z:z^2\to z^2+R^2,
(Edisk)z=ηz4ϵ0∫z2z2+R2u−3/2du=ηz4ϵ0[−2u−1/2]z2z2+R2=ηz2ϵ0[1z−1z2+R2](E_\text{disk})_z=\frac{\eta z}{4\epsilon_0}\int_{z^2}^{z^2+R^2}u^{-3/2}du=\frac{\eta z}{4\epsilon_0}\left[-2u^{-1/2}\right]_{z^2}^{z^2+R^2}=\frac{\eta z}{2\epsilon_0}\left[\frac{1}{z}-\frac{1}{\sqrt{z^2+R^2}}\right]
Mutliplying through by zz,
(Edisk)z=η2ϵ0[1−zz2+R2](E_\text{disk})_z=\frac{\eta}{2\epsilon_0}\left[1-\frac{z}{\sqrt{z^2+R^2}}\right]
This is only valid for z>0z>0. If z<0z<0, the magnitude is the same in the opposite direction.

When z≫Rz\gg R, we factor out z2z^2 from the denominator to get
η2ϵ0[1−11+R2/z2]=η2ϵ0(1−(1+R2/z2)−1/2)\frac{\eta}{2\epsilon_0}\left[1-\frac{1}{\sqrt{1+R^2/z^2}}\right]=\frac{\eta}{2\epsilon_0}(1-(1+R^2/z^2)^{-1/2})
When x≪1x\ll1, the binomial approximation (1+x)n≈1+nx(1+x)^n\approx1+nx can be used. Using it gives
η2ϵ0(1−(1+(−1/2)(R2/z2)))=η2ϵ0R22z2=Q/πR24ϵ0R2z2=14πϵ0Qz2\frac{\eta}{2\epsilon_0}(1-(1+(-1/2)(R^2/z^2)))=\frac{\eta}{2\epsilon_0}\frac{R^2}{2z^2}=\frac{Q/\pi R^2}{4\epsilon_0}\frac{R^2}{z^2}=\frac{1}{4\pi\epsilon_0}\frac{Q}{z^2}
describes the field of a point charge, as expected.

A Plane of Charge

If the distance zz is small, we can approximate with an infinite plane of charge
From the equation above, we let R→∞R\to\infty, then Eplane=η2ϵ0E_\text{plane}=\frac{\eta}{2\epsilon_0}
Notice that this is dependent only on η\eta, the charge density; in particular, it is not dependent on zz. Thus, the field strength is the same everywhere.

The approximation is valid only if zz is much smaller than RR, so if that is the case, we can think of the field as a constant η/2ϵ0\eta/2\epsilon_0 regardless of location or distance.

We do have to consider direction.
E⃗plane=(∣η∣2ϵ0,{away from plane if charge is +toward the plane if charge is −)\vec{E}_\text{plane}=\left(\frac{|\eta|}{2\epsilon_0},\begin{cases}\text{away from plane if charge is }+\\\text{toward the plane if charge is }-\end{cases}\right)

Sphere of Charge

The integration is skipped, and will be derived later.
The field on a point outside a sphere of charge QQ and radius RR is the same as that of a point charge QQ located in the center
E⃗sphere=Q4πϵ0r2r⃗ for r≥R\vec{E}_\text{sphere}=\frac{Q}{4\pi\epsilon_0r^2}\vec{r}\text{ for }r\ge R

Parallel-Plate Capacitor

Two electrodes with equal and opposite charge arranged face-to-face a distance of dd apart are called parallel-plate capacitors
Assume dd is much smaller than the radius

Capacitors are charged by transferring electrons from one plate to another. The plate that gains NN electrons has charge −Q=N(−e)-Q=N(-e)

From the infinite plane field, we see that outside the capacitor, the fields cancel out. Inside, E⃗+\vec{E}_+ and E⃗−\vec{E}_- point in the smae direction, so the field is large.
Thus, the net field inside is
E⃗capacitor=Qϵ0A, from positive to negative\vec{E}_\text{capacitor}=\frac{Q}{\epsilon_0A},\text{ from positive to negative}
and 0⃗\vec{0} outside.

In an ideal capacitor, the field is a uniform electric field as described above. In reality, there is a weak field outside the capacitor called the fringe field

Motion of Charged Particles

Force on particle qq in a field EE is Fon q=qE⃗=ma⃗F_{\text{on }q}=q\vec{E}=m\vec{a}, so the acceleration is given by
a⃗=qmE⃗\vec{a}=\frac{q}{m}\vec{E}
This ratio, q/mq/m, is the charge-to-mass ratio, which determines the acceleration
In a uniform electric field, E⃗\vec{E} is constant, so acceleration is constant.

Dipole

A field affects a dipole by applying a F⃗+=+qE⃗\vec{F}_+=+q\vec{E} force on the positive end and F⃗−=−qE⃗\vec{F}_-=-q\vec{E} on the negative end. These cancel out to achieve a net force of F⃗net=0⃗\vec{F}_{\text{net}}=\vec{0}. However, a torque is applied instead.

The torque is applied until equilibrium, when the dipole is parallel to the field. →\rightarrow become polarized, leading to an excess of charge on one end
The magnitude of the torque is τ=2×dF⃗+=2(12ssin⁡θ)(qE)=pEsin⁡θ\tau=2\times d\vec{F}_+=2(\frac{1}{2}s\sin\theta)(qE)=pE\sin\theta
When torque is aligned with the field, θ=0\theta=0 and τ=0\tau=0
In vector form,
τ⃗=p⃗×E⃗\vec{\tau}=\vec{p}\times\vec{E}

In a nonuniform field once equilibrium is reached, the force on one end is slightly stronger, creating a net force, always toward the charge.