Electrostatics Concept Page - 12

Example

Potential energy of a charge in external electric field

For a region where electric field exists and electric potential is given by V$V$, electric potential energy of a point charge Q$Q$ is given by U=QV$U=QV$. 
Example:
A body of mass 1kg$1kg$ and charge 1C$1C$ is accelerated from rest by a potential difference of 1V$1V$. Find its final velocity.
Solution:
Change in potential energy ΔU=QΔV=−1J$\mathrm{\Delta }U=Q\mathrm{\Delta }V=-1J$ as body moves from higher to lower potential, ΔV=−1V$\mathrm{\Delta }V=-1V$
From law of conservation of energy, change in kinetic energy ΔK=−ΔU$\mathrm{\Delta }K=-\mathrm{\Delta }U$
12mv2=1${\displaystyle \frac{1}{2}}m{v}^2=1$
v=2√ m/s$v=\sqrt{2}m/s$
Note:
Potential is always defined with respect to a reference and potential energy indicates the energy in bringing the charge to a point from zero potential slowly.

Example

Potential energy of a system of two charges in an external field

Example:
Electric field in a region is given as E⃗ =2i^$\overrightarrow{E}=2\hat{i}$. A charge of 1$1$ C is placed at x=1m$x=1m$ and a charge of 2$2$ C is placed at x=2 m$x=2m$. Find the total energy of the system of charges.
Assume potential at x=0$x=0$ is zero.
Solution:
V=−∫E⃗ .d⃗ x=−2x$V=-\int \overrightarrow{E}.\overrightarrow{d}x=-2x$
At x=1,V1=−2$x=1,V_1=-2$
At x=2,V2=−4$x=2,V_2=-4$
Potential energy of system, U=q1q24πεod+q1V1+q2V2$U={\displaystyle \frac{q_1{q}_2}{4\pi {\epsilon }_{o}d}}+q_1{V}_1+q_2{V}_2$
Substituting values, U=24πεo−10$U={\displaystyle \frac{2}{4\pi {\epsilon }_o}}-10$

Formula

Electric potential energy in creating a charged spherical shell

U=Q28πϵ0R$U={\displaystyle \frac{Q^2}{8\pi {ϵ}_{0}R}}$
where R is the radius of a uniformly charged conducting shell of charge Q.

Formula

Energy in creating a charged spherical sphere

U=3Q220πϵ0R$U={\displaystyle \frac{3Q^2}{20\pi {ϵ}_{0}R}}$
where R is the radius of a uniformly charged sphere of charge Q and constant charge density ρ=3Q4πR3$\rho ={\displaystyle \frac{3Q}{4\pi R^3}}$

Example

Potential energy of a system of charges

Q. Charges +q$+q$, −4q$-4q$ and +2q$+2q$ are arranged at the corners of an equilateral triangle of side 0.15 m$0.15m$. If q=1 μC$q=1\mu C$. Find the potential energy of the system.
Sol : The potential of the system is given by,
U=kq1q2r+kq2q3r+kq3q1r$U={\displaystyle \frac{k{q}_1{q}_2}{r}}+{\displaystyle \frac{k{q}_2{q}_3}{r}}+{\displaystyle \frac{k{q}_3{q}_1}{r}}$
U=9×109×10015×(−4q2−8q2+2q2)$U=9\times {10}^9\times {\displaystyle \frac{100}{15}}\times (-4q^2-8q^2+2q^2)$
U=35×1011×(−10)×10−12=−0.6 J$U={\displaystyle \frac{3}{5}}\times {10}^{11}\times (-10)\times {10}^{-12}=-0.6J$

Example

Problem based on kinetic energy of a body accelerated by an electric potential

Example:
An electron is accelerated through a potential difference of 200 volts. If e/m for the electron be 1.6×1011$1.6\times {10}^{11}$ coulomb/kg, find the velocity acquired by the electron.
Solution:As the electron moves through the potential difference, the change in potential energy will be equal to the kinetic energy gained by the electron.
12×m×v2=e×V${\displaystyle \frac{1}{2}}\times m\times v^2=e\times V$
⇒v2=2×e×Vm$\Rightarrow v^2={\displaystyle \frac{2\times e\times V}{m}}$
v2=2×1.6×1011×200$v^2=2\times 1.6\times {10}^{11}\times 200$
⇒v=8×106m/s$\Rightarrow v=8\times {10}^{6}m/s$

Definition

Electric dipole

An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance 2a. The direction from q to -q is said to be the direction of the dipole.
p=q×2a$p=q\times 2a$
where p$p$ is the electric dipole moment pointing from the negative charge to the positive charge.

Example

Force on electric dipole

A small electric dipole having dipole moment p⃗ $\overrightarrow{p}$ is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod  placed along x axis, with one end coinciding with origin. If linear charge density rod is +λ$+\lambda$ and distance of dipole from rod is ′a′${}^{\prime }{a}^{\prime }$, then calculate the electric force acting on dipole.Electric field at A due to dipole,   EA=2KPr3$E_A={\displaystyle \frac{2KP}{r^3}}$            where K=14πϵo$K={\displaystyle \frac{1}{4\pi {ϵ}_o}}$
EA=2KP(x+a)3$E_A={\displaystyle \frac{2KP}{(x+a{)}^3}}$
Force at A due to dipole,   FA=qEA=(λ$F_A=q{E}_A=(\lambda$ dx)$dx)$ ×2KP(x+a)3$\times {\displaystyle \frac{2KP}{(x+a{)}^3}}$
FA=2λKP(x+a)3dx$F_A={\displaystyle \frac{2\lambda KP}{(x+a{)}^3}}dx$
Total force exerted due to dipole on the rod,    FD=∫∞02λKP(x+a)3dx$F_D={\int }_0^{\mathrm{\infty }}{\displaystyle \frac{2\lambda KP}{(x+a{)}^3}}dx$
FD=2λKP×1−2(x+a)2∣∣∣∞0$F_D=2\lambda KP\times {\displaystyle \frac{1}{-2(x+a{)}^2}}{|}_0^{\mathrm{\infty }}$  ⟹FD=λKPa2$⟹F_D={\displaystyle \frac{\lambda KP}{a^2}}$
Now force exerted by rod on dipole is equal and opposite to FD$F_D$  (Newton's law)
FR→=−FD=−λP4πϵoa2$\overrightarrow{F_R}=-F_D={\displaystyle \frac{-\lambda P}{4\pi {ϵ}_o{a}^2}}$

Definition

Physical significance of dipoles

In most molecules, the centres of positive charges and of negative charges lie at the same place. Therefore, their dipole moment is zero. CO2$C{O}_2$ and CH4$C{H}_4$ are of this type of molecules. However, they develop a dipole moment when an electric field is applied. But in some molecules, the centres of negative charges and of positive charges do not coincide. Therefore they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, H2O$H_{2}O$,is an example of this type. Various materials give rise to interesting properties and important applications in the presence or absence of electric field.

Example

Electric field of an electric dipole for axial points

Two charges 10μC$10\mu C$ are placed 5.0mm$5.0mm$ apart.Determine the electric field at a point P on the axis of the dipole 15 cm$15cm$ away from its centre O on the side of the positive charge.
Solution:
Field at P due to charge +10μC$+10\mu C$:
=10−54π(8.854×10−12C2N−1m−2)×1(15−0.25)2×10−4m2$={\displaystyle \frac{{10}^{-5}}{4\pi (8.854\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2})}}\times {\displaystyle \frac{1}{(15-0.25{)}^2\times {10}^{-4}{m}^2}}$
=4.13×106NC−1$=4.13\times {10}^{6}N{C}^{-1}$ along BP
Field at P due to charge −10μC$-10\mu C$
=10−54π(8.854×10−12C2N−1m−2)×1(15+0.25)2×10−4m2$={\displaystyle \frac{{10}^{-5}}{4\pi (8.854\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2})}}\times {\displaystyle \frac{1}{(15+0.25{)}^2\times {10}^{-4}{m}^2}}$
=3.86×106NC−1$=3.86\times {10}^{6}N{C}^{-1}$ along PA
The resultant electric field at P due to the two charges at A and B is
=2.7×105NC−1$=2.7\times {10}^{5}N{C}^{-1}$ along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole.For a dipole consisting of charges q,2a$q,2a$ distance apart, the electric field at a distance r$r$ from the centre on the axis of the dipole has a magnitude
E=2p4πϵ0r3(r/a>>1)$E={\displaystyle \frac{2p}{4\pi {ϵ}_0{r}^3}}(r/a>>1)$
where p=2aq$p=2aq$ is the magnitude of the dipole moment.The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from −q to q$-qtoq$). Here,
p=10−5C×5×10−3m=5×10−8Cm$p={10}^{-5}C\times 5\times {10}^{-3}m=5\times {10}^{-8}Cm$
E=2.6×105NC−1$E=2.6\times {10}^{5}N{C}^{-1}$Along the dipole moment direction AB, which is close to the result obtained earlier.

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