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Electric Field

The electric field created by a charge, QQ, acting on a charge qq, is given by:

E=Feq=ke∣Q∣r2.E = \frac{F_e}{q} = k_e\frac{|Q|}{r^2}.

Or in vector form:

E=Feq=keQr2r^,\boldsymbol{E} = \frac{\boldsymbol{F_e}}{q} = k_e\frac{Q}{r^2} \boldsymbol{\hat{r}},

where:

r=∣rq−rQ∣,r^=rq−rQr, \begin{align*} r &= |\boldsymbol{r}_q - \boldsymbol{r}_Q|, \\ \boldsymbol{\hat{r}} &= \frac{\boldsymbol{r}_q - \boldsymbol{r}_Q}{r}, \end{align*}

where rQ\boldsymbol{r}_Q is the position of the source charge and rq\boldsymbol{r}_q is the position of the charge that the electric field acts on.

The unit of electric field is newton per coulomb (NC−1NC^{-1}).

A 10 nC10\ nC charge is placed at the origin. A second −20 nC-20\ nC charge is placed 50 cm50\ cm to the right of the first charge. What is the magnitude and direction of the electric field 20 cm20\ cm above the first charge?

Q1=10−8 C,Q2=−2⋅10−8 C,r1=0,r2=0.5 m x^,rt=0.2 m y^,rt1=0.2 m y^,rt1=0.2 m,r^t1=y^,rt2=−0.5 m x^+0.2 m y^,rt2=2910 m,r^t2=−52929 x^+22929 y^,E1=keQ1rt12r^t1=2250 NC−1 y^,E2=keQ2rt22r^t2=−1800029 NC−1 r^t2=9000029841 NC−1 x^−3600029841 NC−1 y^,≈576.3 NC−1 x^−230.52 NC−1 y^,E=E1+E2=9000029841 NC−1 x^+(2250−3600029841) NC−1 y^,≈576.3 NC−1 x^+2019.48 NC−1 y^,E≈2100.1 NC−1. \begin{align*} Q_1 &= 10^{-8}\ C, \\ Q_2 &= -2 \cdot 10^{-8}\ C, \\ \boldsymbol{r}_1 &= \boldsymbol{0}, \\ \boldsymbol{r}_2 &= 0.5\ m\ \boldsymbol{\hat{x}}, \\ \boldsymbol{r}_t &= 0.2\ m\ \boldsymbol{\hat{y}}, \\[1.5ex] \boldsymbol{r}_{t1} &= 0.2\ m\ \boldsymbol{\hat{y}}, \\ r_{t1} &= 0.2\ m, \\ \boldsymbol{\hat{r}_{t1}} &= \boldsymbol{\hat{y}}, \\[1.5ex] \boldsymbol{r}_{t2} &= -0.5\ m\ \boldsymbol{\hat{x}} + 0.2\ m\ \boldsymbol{\hat{y}}, \\ r_{t2} &= \frac{\sqrt{29}}{10}\ m, \\ \boldsymbol{\hat{r}_{t2}} &= -\frac{5 \sqrt{29}}{29}\ \boldsymbol{\hat{x}} + \frac{2 \sqrt{29}}{29}\ \boldsymbol{\hat{y}}, \\[1.5ex] \boldsymbol{E}_1 &= k_e \frac{Q_1}{r_{t1}^2} \boldsymbol{\hat{r}_{t1}} \\ &= 2250\ NC^{-1}\ \boldsymbol{\hat{y}}, \\ \boldsymbol{E}_2 &= k_e \frac{Q_2}{r_{t2}^2} \boldsymbol{\hat{r}_{t2}} \\ &= -\frac{18000}{29}\ NC^{-1}\ \boldsymbol{\hat{r}_{t2}} \\ &= \frac{90000 \sqrt{29}}{841}\ NC^{-1}\ \boldsymbol{\hat{x}} - \frac{36000 \sqrt{29}}{841}\ NC^{-1}\ \boldsymbol{\hat{y}}, \\ &\approx 576.3\ NC^{-1}\ \boldsymbol{\hat{x}} - 230.52\ NC^{-1}\ \boldsymbol{\hat{y}}, \\ \boldsymbol{E} &= \boldsymbol{E}_1 + \boldsymbol{E}_2 \\ &= \frac{90000 \sqrt{29}}{841}\ NC^{-1}\ \boldsymbol{\hat{x}} + \left(2250 - \frac{36000 \sqrt{29}}{841}\right)\ NC^{-1}\ \boldsymbol{\hat{y}}, \\ &\approx 576.3\ NC^{-1}\ \boldsymbol{\hat{x}} + 2019.48\ NC^{-1}\ \boldsymbol{\hat{y}}, \\ E &\approx 2100.1\ NC^{-1}. \end{align*}

Two opposite charges with magnitude qq are placed at a distance dd from each other. Calculate the electric field at the distance aa from the midpoint. This point is at a distance rr from both charges:

Electric dipole illustration
r=(d2)2+a2=d2+4a24=d2+4a22. \begin{align*} r &= \sqrt{\left(\frac{d}{2}\right)^2 + a^2} \\ &= \sqrt{\frac{d^2 + 4a^2}{4}} \\ &= \frac{\sqrt{d^2 + 4a^2}}{2}. \\ \end{align*}

Let r1\boldsymbol{r_1} be the position of the first charge, r2\boldsymbol{r_2} the position of the second charge and rt\boldsymbol{r_t} the at which the electric field is calculated:

r=∣rt−r1∣=∣rt−r2∣,d=∣r1−r2∣,r^s1=rt−r1r,r^s2=rt−r2r. \begin{align*} r &= |\boldsymbol{r_t} - \boldsymbol{r_1}| = |\boldsymbol{r_t} - \boldsymbol{r_2}|, \\ d &= |\boldsymbol{r_1} - \boldsymbol{r_2}|, \\ \boldsymbol{\hat{r}_{s1}} &= \frac{\boldsymbol{r}_t - \boldsymbol{r}_1}{r}, \\ \boldsymbol{\hat{r}_{s2}} &= \frac{\boldsymbol{r}_t - \boldsymbol{r}_2}{r}. \end{align*}

The electric field due to the first charge, E1\boldsymbol{E}_1 is equal to:

E1=keqr2r^s1=keqr3(rt−r1). \begin{align*} \boldsymbol{E}_1 &= k_e \frac{q}{r^2} \boldsymbol{\hat{r}_{s1}} \\ &= k_e \frac{q}{r^3} (\boldsymbol{r}_t - \boldsymbol{r}_1). \end{align*}

The electric field due to the second charge, E2\boldsymbol{E}_2 is equal to:

E2=−keqr2r^s2=−keqr3(rt−r2). \begin{align*} \boldsymbol{E}_2 &= -k_e \frac{q}{r^2} \boldsymbol{\hat{r}_{s2}} \\ &= -k_e \frac{q}{r^3} (\boldsymbol{r}_t - \boldsymbol{r}_2). \end{align*}

The total electric field, E\boldsymbol{E} is equal to:

E=E1+E2=keqr3(rt−r1)−keqr3(rt−r2)=keqr3(rt−r1−rt+r2)=keqr3(r2−r1)=keq(d2+4a22)3(r2−r1)=8keq(d2+4a2)32(r2−r1). \begin{align*} \boldsymbol{E} &= \boldsymbol{E}_1 + \boldsymbol{E}_2 \\ &= k_e \frac{q}{r^3} (\boldsymbol{r}_t - \boldsymbol{r}_1) - k_e \frac{q}{r^3} (\boldsymbol{r}_t - \boldsymbol{r}_2) \\ &= k_e \frac{q}{r^3} (\boldsymbol{r}_t - \boldsymbol{r}_1 - \boldsymbol{r}_t + \boldsymbol{r}_2) \\ &= k_e \frac{q}{r^3} (\boldsymbol{r}_2 - \boldsymbol{r}_1) \\ &= k_e \frac{q}{\left(\frac{\sqrt{d^2 + 4a^2}}{2}\right)^3} (\boldsymbol{r}_2 - \boldsymbol{r}_1) \\ &= \frac{8 k_e q}{(d^2 + 4a^2)^{\frac{3}{2}}} (\boldsymbol{r}_2 - \boldsymbol{r}_1). \end{align*}

Transforming the coordinate system so that r1\boldsymbol{r}_1 is at the origin and r2\boldsymbol{r}_2 lies on the x axis, the magnitude of the electric field can be calculated in the original terms:

r2=dx^,E=8keq(d2+4a2)32dx^=8keqd(d2+4a2)32x^,E=8keqd(d2+4a2)32. \begin{align*} \boldsymbol{r}_2 &= d \boldsymbol{\hat{x}}, \\ \boldsymbol{E} &= \frac{8 k_e q}{(d^2 + 4a^2)^{\frac{3}{2}}} d \boldsymbol{\hat{x}} \\ &= \frac{8 k_e q d}{(d^2 + 4a^2)^{\frac{3}{2}}} \boldsymbol{\hat{x}}, \\ E &= \frac{8 k_e q d}{(d^2 + 4a^2)^{\frac{3}{2}}}. \\ \end{align*}

A line has a constant linear charge density λ\lambda. Its ends lie in points AA and BB. What is the electric field at point CC?

Line charge illustration

Let HH be the projection of CC onto the line, then:

Line charge illustration 2
l=B−A,h=H−A,s=C−H, \begin{align*} \boldsymbol{l} &= B - A, \\ \boldsymbol{h} &= H - A, \\ \boldsymbol{s} &= C - H, \\ \end{align*}

Let's transform the coordinate system so that HH lies in origin and the line is aligned with the y-axis:

x^=ss,y^=ll,hy=y^⋅(C−A),h=hyy^,H=A+h. \begin{align*} \boldsymbol{\hat{x}} &= \frac{\boldsymbol{s}}{s}, \\ \boldsymbol{\hat{y}} &= \frac{\boldsymbol{l}}{l}, \\ h_y &= \boldsymbol{\hat{y}} \cdot (C - A), \\ \boldsymbol{h} &= h_y \boldsymbol{\hat{y}}, \\ H &= A + \boldsymbol{h}. \end{align*}

Let's parametrize the position on the line by yy:

Line charge illustration 3
λ=dqdl  ⟹  dq=λ dl=λ dy,r=y2+s2,dE=kedqr2=keλdyy2+s2,dEx=cos⁡α1 dE,dEy=sin⁡α1 dE,cos⁡α1=sr,sin⁡α1=−sin⁡α2=−yr,dEx=keλsdy(y2+s2)32,Ex=keλs∫−hyl−hydy(y2+s2)32=keλs[yy2+s2]−hyl−hy=keλs[l−hy(l−hy)2+s2+hyhy2+s2],dEy=keλy dy(y2+s2)32,Ey=−keλ∫−hyl−hyy dy(y2+s2)32=keλ[1y2+s2]−hyl−hy=keλ[1y2+s2]−hyl−hy=keλ[1(l−hy)2+s2−1hy2+s2]. \begin{align*} \lambda &= \frac{dq}{dl} \implies dq = \lambda\ dl = \lambda\ dy, \\ r &= \sqrt{y^2 + s^2}, \\ dE &= k_e \frac{dq}{r^2} \\ &= k_e \lambda \frac{dy}{y^2 + s^2}, \\ d E_x &= \cos \alpha_1\ dE, \\ d E_y &= \sin \alpha_1\ dE, \\ \cos \alpha_1 &= \frac{s}{r}, \\ \sin \alpha_1 &= -\sin \alpha_2 = -\frac{y}{r}, \\ d E_x &= k_e \lambda s \frac{dy}{(y^2 + s^2)^{\frac{3}{2}}}, \\ E_x &= k_e \lambda s \int_{-h_y}^{l - h_y} \frac{dy}{(y^2 + s^2)^{\frac{3}{2}}} \\ &= \frac{k_e \lambda}{s} \left[\frac{y}{\sqrt{y^2 + s^2}}\right]_{-h_y}^{l - h_y} \\ &= \frac{k_e \lambda}{s} \left[\frac{l - h_y}{\sqrt{(l - h_y)^2 + s^2}} + \frac{h_y}{\sqrt{h_y^2 + s^2}}\right], \\ d E_y &= k_e \lambda \frac{y\ dy}{(y^2 + s^2)^{\frac{3}{2}}}, \\ E_y &= -k_e \lambda \int_{-h_y}^{l - h_y} \frac{y\ dy}{(y^2 + s^2)^{\frac{3}{2}}} \\ &= k_e \lambda \left[\frac{1}{\sqrt{y^2 + s^2}}\right]_{-h_y}^{l - h_y} \\ &= k_e \lambda \left[\frac{1}{\sqrt{y^2 + s^2}}\right]_{-h_y}^{l - h_y} \\ &= k_e \lambda \left[\frac{1}{\sqrt{(l - h_y)^2 + s^2}} - \frac{1}{\sqrt{h_y^2 + s^2}}\right]. \\ \end{align*}

Putting it all together, the electric field vector is equal to:

E=Exx^+Eyy^\boldsymbol{E} = E_x \boldsymbol{\hat{x}} + E_y \boldsymbol{\hat{y}}

where:

l=B−A,hy=y^⋅(C−A),h=hyy^,H=A+h,s=C−H,Ex=keλs(l−hy(l−hy)2+s2+hyhy2+s2),Ey=keλ(1(l−hy)2+s2−1hy2+s2),x^=ss,y^=ll, \begin{align*} \boldsymbol{l} &= B - A, \\ h_y &= \boldsymbol{\hat{y}} \cdot (C - A), \\ \boldsymbol{h} &= h_y \boldsymbol{\hat{y}}, \\ H &= A + \boldsymbol{h}, \\ \boldsymbol{s} &= C - H, \\ E_x &= \frac{k_e \lambda}{s} \left(\frac{l - h_y}{\sqrt{(l - h_y)^2 + s^2}} + \frac{h_y}{\sqrt{h_y^2 + s^2}}\right), \\ E_y &= k_e \lambda \left(\frac{1}{\sqrt{(l - h_y)^2 + s^2}} - \frac{1}{\sqrt{h_y^2 + s^2}}\right), \\ \boldsymbol{\hat{x}} &= \frac{\boldsymbol{s}}{s}, \\ \boldsymbol{\hat{y}} &= \frac{\boldsymbol{l}}{l}, \\ \end{align*}

and HH lies at the origin.

Let l→∞l \to \infty and hy=l2h_y = \frac{l}{2}. Since the point CC would be at the center of the line, the yy component of the electric field vanishes:

Ey=keλ(1(l−l2)2+s2−1(l2)2+s2)=keλ(1(l2)2+s2−1(l2)2+s2)=0. \begin{align*} E_y &= k_e \lambda \left(\frac{1}{\sqrt{\left(l - \frac{l}{2}\right)^2 + s^2}} - \frac{1}{\sqrt{\left(\frac{l}{2}\right)^2 + s^2}}\right) \\ &= k_e \lambda \left(\frac{1}{\sqrt{\left(\frac{l}{2}\right)^2 + s^2}} - \frac{1}{\sqrt{\left(\frac{l}{2}\right)^2 + s^2}}\right) \\ &= 0. \end{align*}

Taking the limit of ExE_x:

Ex=keλslim⁡l→∞(l−l2(l−l2)2+s2+l2(l2)2+s2)=keλslim⁡l→∞(l2(l2)2+s2+l2(l2)2+s2)=keλslim⁡l→∞(ll2+4s24)=2keλslim⁡l→∞(11+4s2l2)=2keλs. \begin{align*} E_x &= \frac{k_e \lambda}{s} \lim_{l \to \infty} \left(\frac{l - \frac{l}{2}}{\sqrt{\left(l - \frac{l}{2}\right)^2 + s^2}} + \frac{\frac{l}{2}}{\sqrt{\left(\frac{l}{2}\right)^2 + s^2}}\right) \\ &= \frac{k_e \lambda}{s} \lim_{l \to \infty} \left(\frac{\frac{l}{2}}{\sqrt{\left(\frac{l}{2}\right)^2 + s^2}} + \frac{\frac{l}{2}}{\sqrt{\left(\frac{l}{2}\right)^2 + s^2}}\right) \\ &= \frac{k_e \lambda}{s} \lim_{l \to \infty} \left(\frac{l}{\sqrt{\frac{l^2 + 4s^2}{4}}}\right) \\ &= \frac{2 k_e \lambda}{s} \lim_{l \to \infty} \left(\frac{1}{\sqrt{1 + 4\frac{s^2}{l^2}}}\right) \\ &= \frac{2 k_e \lambda}{s}. \end{align*}