Potential energy of a pair of point charges Q & q is $U=\frac{Qq}{4 \pi \epsilon_0 r}$
Potential energy of a system of charges
$U=\frac{q_1q_2}{4 \pi \epsilon_0 r_{12}}+\frac{q_1q_3}{4 \pi \epsilon_0 r_{13}}+\frac{q_2q_3}{4 \pi \epsilon_0 r_{23}}$
Electrostatics potential for a point charge Q.
$V(r)=\frac{Q}{4 \pi \epsilon_0 r}$
V(r) is a scalar quantity.
$V\vec{(r)}$ and $\vec{E}\vec{(r)}$ are related.
Unit of potential 1 Volt = 1 J/C
Unit of energy : electron volt(eV)
$1eV=(1e)\times 1V=1.6 \times 10^{-19}J$
Work done by an external force in moving a unit charge from a point $r_i$ to $r_f$ is
$W=V(r_f)-V(r_i)$
For a point charge
$W=\frac{Q}{4 \pi \epsilon_0}(\frac{1}{r_f}-\frac{1}{r_i})$
Superposition Principle
$V(r) = \frac{q_1}{4 \pi \epsilon_0 r_1}+\frac{q_2}{4 \pi \epsilon_0 r_2}+\frac{q_3}{4 \pi \epsilon_0 r_3}$
$= \sum\frac{q_i}{4 \pi \epsilon_0 r_i}$
Distribution of charges
Potential of a charged conducting sphere-
$\vec{E}=\frac{Q \hat{r}}{4 \pi \epsilon_0 r^2},r>R$ = 0, r<R
V(r) = $\int_\infty^r$ $\vec{F}_{enc}$ . $d\vec{r}$
$ = -\frac{Q}{4 \pi \epsilon_0}$
$\int_\infty^r \frac{dr}{r^2}$
$= \frac{Q}{4 \pi \epsilon_0 r}$ r>R
$V(R) = \frac{Q}{4 \pi \epsilon_0 R}$ at r=R
Conducting sphere of radius R = 10cm = 0.1m
$Q = 1nC=10^{-9}C$
$V=\frac{Q}{4 \pi \epsilon_0 R}=\frac{10^{-9}\times9\times10^9}{0.1}=90V$
$E=\frac{Q}{4 \pi \epsilon_0 R^2}=\frac{V}{R}=\frac{90}{0.1}=900V/m$
$E_{max}=3\times10^6V/m$ in out
R = 0.1m
Maximum Potential
$V_{max} = E_{max} R$ $=3\times10^6\times0.1=3\times10^5V$ = 300 kV
R = 1cm = 0.01m
$V_{max}=300kV$
$V(P)=\frac{q}{4 \pi \epsilon_0 r_1}-\frac{q}{4 \pi \epsilon_0 r_2}$
$=\frac{q}{4 \pi \epsilon_0}(\frac{1}{r_1}-\frac{1}{r_2})$
$r_1^2 = r^2+a^2-2ar\cos\theta$
$2_2^2=r^2+a^2+2ar\cos\theta$
$r_1^2=r^2+a^2-2ar\cos\theta$ $=r^2(1+\frac{a^2}{r^2}-\frac{2a}{r}\cos\theta)$
$r_1=r(1+\frac{a^2}{r^2}-\frac{2a}{r}\cos\theta)^{1/2}$
$\frac{1}{r_1}=\frac{1}{r}(1+\frac{a^2}{r^2}-\frac{2a}{r}\cos\theta)^{-1/2}$
r » a
$\frac{1}{r_1}\simeq\frac{1}{r}(1+\frac{a}{r}\cos\theta)$
Negative terms of order $\frac{a^2}{r^2}$ and greater.
$r_2^2=r^2+a^2+2ar\cos\theta$
$\frac{1}{r_2}\simeq\frac{1}{r}(1-\frac{a}{r}\cos\theta)$
$\frac{1}{r_1}-\frac{1}{r_2}\simeq\frac{2a}{r^2}\cos\theta$
$V(P)=\frac{q}{4 \pi \epsilon_0}.\frac{2a}{r^2}\cos\theta$
$\vec{|p|}=q.2q$
$V(P)=\frac{\vec{|p|}\cos\theta}{4 \pi \epsilon_0 r^2}=\frac{\vec{p}.\hat{g}}{4 \pi \epsilon_0 r^2}$
$V(r)=\frac{\vec{p}.\hat{r}}{4 \pi \epsilon_0 r^2}$r»a
$V(r)=\frac{P\cos\theta}{4 \pi \epsilon_0 r^2}$
$\theta=0;V(r)=\frac{p}{4 \pi \epsilon_0 r^2}$
$\theta=\pi;V(r)=-\frac{p}{4 \pi \epsilon_0 r^2}$
$\theta=\frac{\pi}{2};V(r)=0$
$2 \pi r.l.E=\frac{\lambda l}{\epsilon_0}$
$\vec{E}=\frac{\lambda}{2\pi\epsilon_0r}.\hat{r}$
$W=-int_{r_a}^{r_b}\frac{\lambda \hat{r}}{2 \pi \epsilon_0 r}.\hat{r}dr$
$=-\frac{\lambda}{2 \pi \epsilon_0}\int_{r_a}^{r_b}\frac{dr}{r}=\frac{\lambda}{2 \pi \epsilon_0}ln(\frac{r_a}{r_b})$
Zero of potential at same r value
Let V=0 at $r=r_a=R$ and let $r_b=r$
$V(r)=\frac{\lambda}{2 \pi \epsilon_0}ln(\frac{R}{r})$
$V=\frac{Q}{4 \pi \epsilon_0 r}$
$r=r_1 ; V=V_1=\frac{Q}{4 \pi \epsilon_0 r_1}$
$r=r_2;v=v_2=\frac{Q}{4 \pi \epsilon_0 r_2}$
$r_2>r_1$
What about if
$v_2>v_1$ and $v_1>v_2?$
Consider two adjacent equipotentials surfaces
Work done by the external force in moving a unit charge from point A to point B
$dW=(v_0+dv)-v_0$ = dv
Work done in also given by $-\vec{E}.\vec{dl}$ $=-Edl\cos \theta$
$Edl\cos\theta=-dv$
$E\cos\theta=-\frac{dv}{dl}$
$E_x=-\frac{dv}{dx}$
$E_y=-\frac{dv}{dy}$
$E_z=-\frac{dv}{dz}$
$=\frac{Q}{4 \pi \epsilon_0(x^2+y^2+z^2)^{1/2}}$
$E_x=-\frac{dv}{dx}=-\frac{Q(-1/2).2x}{4 \pi \epsilon_0 (x^2+y^2+z^2)^{3/2}}$
$=\frac{Q}{4 \pi \epsilon_0 (x^2+y^2+z^2)}\frac{x}{\sqrt{x^2+y^2+z^2}}$
$=\frac{Q}{4 \pi \epsilon_0r^2}(\frac{x}{r})$
An electric dipole of moment $\vec{p}=10\hat{k}nC.m$ is located at the origin in force sphere. Calculate the potential or a point P with coordinates.
$x_P=0.5m$
$y_p=0$
$z_P=0.87m$