Electric Field and Potential

 ELECTRIC FIELD

Electric Field and Electric Field Intensity$\\$

• The space around an electric charge in which its influence can be experienced is known as electric field.$\\$

• The electric field intensity $\vec{E}$ at any point is equal in magnitude to the force experienced per unit (test) position charge placed at that point and is directed along the direction of the force experienced.$\\$i.e $\vec{E}$=$\frac{\vec{F}}{q}$⇒$\vec{F}$=qE$\\$

• Electric field intensity is a vector quantity. Electric field intensity due to positive charge is always away from the charge and due to negative charge is

always towards the charge.$\\$

• Its unit is Newton/Coulomb (N$C^{-1}$)or Volt/meter (V$m^{-1}$ )$\\$
• Electric field due to point charge q at distance r is: E=$\frac{1}{4πε_0}\frac{q}{r^2} \\$
• $\vec{E}$=$\vec{E_1}$+$\vec{E_2}$+$\vec{E_3}$+ ……………$\\$
• The magnitude of the resultant of two electric fields is given by E=$\sqrt{E_1 ^2+E_2 ^2+2E_1 E_2 cos⁡θ} \\$

Electric Lines of Force

The line or a curve along which an isolated +ve charge would travel if it is free to move in an electric field is known as electric line of force.

They start from positively charged body and end at a negatively charged body.

No electric lines of force exist inside the charged body.

Tangent to the line of force at any point gives the direction of electric intensity at that point.

No two electric lines of force can intersect each other.

The electric lines of force are always normal to the surface of a conductor, both while starting and ending on the conductor. Therefore, there is no component of electric field intensity parallel to the surface of the conductor.

They never form closed loops.

They are always perpendicular to equipotential surface.

The electric lines of force contract longitudinally, on account of attraction between unlike charges.

The electric lines of force exert a lateral pressure on account of repulsion between like charges.

In uniform electric field, the electric lines of force are equidistant, parallel straight lines.

When a metallic solid sphere is placed in a uniform electric field, then the lines of force are normal to the surface at every point but they cannot pass through the conductor.

Field Intensity in Special Cases

Intensity of the electric field inside a charge spherical conductor/ hollow/ spherical shell/ conducting sphere is zero, since charge resides on the outer surface of the conductor. But $E_{surface}=\frac{1}{4πε_o} \frac{q}{R^2}$ (Where R= radius of the sphere) $E_{outside} =\frac{1}{4πε_o} \frac{q}{r^2}$ (For r>R )

The intensity of electric field for a uniformly charged non- conducting sphere of radius R is given by$\\$$E_{outside} =\frac{1}{4πε_o} \frac{q}{r^2}$ (For r>R )$\\$

$E_{surface}=\frac{1}{4πε_o}.\frac{q}{R^2}$ (Where r=R is radius of the sphere)$\\$$E_{inside} =\frac{1}{4πε_o}. \frac{qr}{R^3}$ ( For r<R)$\\$$E_{centre}$ = 0 ( for r=0)$\\$

Intensity of electric field at some point on the axis of uniformly charged ring of radius R is given by$\\$E=$\frac{1}{4πε_o}\frac{qx}{(x^2+R^2 )^{3/2}} =\frac{λRx}{2ε_o (x^2+R^2 )^{3/2}}$$\\$But it is zero at the center of ring.$\\$

Intensity of electric field near an infinite rod of charge is given by$\\$E=$\frac{λ}{2πε_0 r}$$\\$where λ is the linear charge density and r is the distance from the axis of rod.

Intensity of the electric field near a non-conducting infinite sheet of charge is E=$\frac{σ}{2ε_o}$$\\$where σ is the surface charge density.

Electric Dipole

INTRO:$\\$

• Two equal and opposite charges

separated by a small distance constitute an electric dipole.$\\$

• Electric dipole moment is a vector

quantity (p) whose magnitude is equal to the product of magnitude of one charge and the distance between the two charges.$\\$

Do You Know?$\\$Dipole moment is a vector whose direction is from negative charge (-q) to positive charge (+q).$\\$Electric dipole moment$\\$$\vet{p}$=q.d$∴$\vet{p}$=q(2l)

Electric field intensity on equatorial/ broadside on position/ Tan B position:$\\$Magnitude of electric field produced by dipole,$\\$$E_b=\frac{p}{4πε_o (r^2 + l^2 )^{3/2}}$$\\$For short dipole, r> > >d$\\$$E_b=\frac{1}{4πε_0}. \frac{p}{r^3}$ ⇒ $E_b∝\frac{1}{r^3}$$\\$

Electric field intensity on axial/end on/tan A position:$\\$Electric field produced by dipole at end on position is:$\\$$E_a=\frac{1}{4πε_0}. \frac{2pr}{(r^2-l^2 )^2}$$\\$For short dipole, r >>> d , so$\\$$E_a=\frac{1}{4πε_0}. \frac{2p}{r^3}$ ⇒ $E_a∝\frac{1}{r^3}$$\\$From 1 and 2,$\\$$E_{axial} = 2 E_{broad}$

Do You Know ?$\\$

• Angle between $\vec{p}$ and $\vec{E}$ on broad on position is $180^o$ and it is $0^o$ at end on position.
• Equatorial line is also called equipotential surface or zero potential surfaces.
• Electric potential due to dipole at end on position is V=$\frac{p}{4πεₒr^3}$

Electric field at any point around the dipole:$\\$E=$\frac{p}{4πεₒr^2 }\sqrt{1+3 cos^2⁡θ}$$\\$Where θ is the angle made by position vector $\vec{r}$ with dipole moment $\vec{p} \\$[Note: V=$\frac{p cosθ}{4πEₒr^2 }$ → ∝ $\frac{1}{r^2}$ In case of dipole]$\\$Electrostatic force between two short dipoles of dipole moments $p_1 and p_2$ at separation r is $F_1=\frac{1}{4πε_0}⋅\frac{6p_1.p_2}{r^4}$ ( when coaxial)$\\$$F_2=\frac{1}{4πε_o}⋅\frac{3p_1.p_2}{r^4}$$\\$(when mutually perpendicular)$\\$∴$F_1=2F_2⇒F∝\frac{1}{r^4}$$\\$But, F∝$\frac{1}{r^3}$ for a dipole.

Dipole kept in uniform electric field:$\\$If θ be the angle between the direction of uniform electric field E and axis of dipole,$\\$then torque acting on the dipole is τ=$\vec{p}×\vec{E} \\$τ=pEsin⁡θ$\\$[Where θ is the angle between p and E ]

Do You Know?$\\$Net force $F_{net}$ acting on dipole is zero i.e. $F_{net}$ = 0 Hence, there is no linear or transitory motion. Dipole only rotates.$\\$Case I: dipole kept normal to electric field i.e.θ =$90^o \\$$F_{net}$ = 0 (always)$\\$$τ_{max}$ = pE$\\$Case II: Dipole kept along electric field ie θ=$0^o or 180^0 \\$$F_{net}$ = 0 (always)$\\$$τ_{min}$ = 0$\\$

• When p is along $\vec{E}$ i.e θ=$0^o \\$

$τ_{min}$ = 0$\\$The dipole is in stable equilibrium (minimum P.E)$\\$

• When dipole is held opposite direction to $\vec{E}$ i.e θ=$180^o \\$

$τ_{min}$ = 0$\\$the dipole is in unstable equilibrium. (maximum P.E)$\\$Thus, as θ increases from $0^o$ to $0^o$, r first increases, becomes maximum at $90^o$ and then again decreases to zero.

Work done in revolving a dipole$\\$

• Work done in revolving a dipole from initial angle $θ_1$ to final angle $θ_2$ against the electric field is given by,$\\$

W=pE(cos⁡$θ_1$-cos⁡$θ_2$ )$\\$Where $θ_1$ is initial angle between $\vec{p}$ and $\vec{E}$ and $θ_2$ is final angle between $\vec{p}$ and $\vec{E}$.$\\$If initial angle is not given then assume θ=$0^o$, since external agent do the work always from stable equilibrium state of the body. $W_{ext}$ =pE⁡(1-cos⁡θ)$\\$

• The work done to separate the charges of electric dipole to infinity is known as binding energy of the dipole.$\\$

$U_B=\frac{q^2}{4πε_0 (2l)}$$\\$

Electrostatic potential energy (U) of a dipole in uniform electric field $\vec{E} \\$U=-$\vec{p}⋅\vec{E}$$\\$

∴U=-pEcos⁡θ$\\$

(i) For θ=$0^o$, U=-pE. It is the minimum value of potential energy and the dipole is in stable equilibrium.$\\$

(ii) For θ=$90^o$ , U=0$\\$

(iii) For θ=$180^o$, $U_{max}$ = pE and the dipole is in unstable equilibrium.

Electric flux (ϕ)

It is defined as the product of electric intensity and area provided area is kept normal to electric intensity.$\\$ϕ=$\vec{E}⋅\vec{A } \\$∴ϕ=E⋅Acos⁡θ$\\$where θ is the angle between E and normal on area.$\\$

The flux can be positive, if field lines leave the area, negative if field lines enter the area and zero if field line parallel to plane i.e. area vector is normal to electric field.

It is a scalar quantity and its unit is Volt -meter (Vm) / N$m^2 C^{-1}$

Gauss’s theorem$\\$It states that the total electric flux coming out of a closed body is equal to $\farc{1}{ε_0}$ times the net charge enclosed by the body.$\\$

$ϕ_{net}=\frac{q_{net}}{ε_0}$

Gauss’s theorem is applicable for closed surface only.

Gaussian surface should always be closed.

Even if total net flux through a closed surface is zero, the electric field E at the Gaussian surface may be non zero

Total flux from cube with charge + q placed at the centre is $ϕ=\frac{q}{ε_0}$ Flux through one face =$\frac{q/ε_0}{6}$

ϕ is independent of the radius of the closed surface.

Coulomb’s law can be deducted from Gauss’s Theorem

Force per unit area (or electric pressure) on a charged conductor$\\$

If σ is the surface charge density, electric pressure$\\$$\frac{dF}{dA}$=($σ^2/2 g_e$ )=$\frac{1}{2} ε_0 E^2 \\$

The energy stored per unit volume in electric field i.e. energy density of electric field is$\\$$\frac{σ^2}{2ε_0}=\farc{1}{2}ε_0 E^2 \\$

If $ε_r$ is the relative permittivity, then energy density =$\frac{1}{2} ε_r ε_0 E^2 \\$

The force is always directed outward as $(±σ)^2$ is positive i.e. whether body is charged +ve ly or -ve ly, this force will always try to expand the charged body.

A soap bubble or rubber balloon expands on giving a +ve or -ve charge to it.

Electric Potential

Electric potential at a point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point.

Potential difference between two points A and B in an electric field is the work required to more a unit positive charge from the point A to the point B against the electric field.$\\$$\fracV_B - V_A= \frac{W_{AB}}{q_0}$$\\$

$V_{AB} = \frac{W_{AB}}{q_0}$$\\$(i) If $W_{AB}$ is positive, $V_B > V_A \\$(ii) If $W_{AB}$ is negative, $V_B < V_A \\$(iii) If $W_{AB}$ is zero, $V_B = V_A \\$

The potential at a point due to positive charge is positive while due to negative charge, it is negative.

When a positive charge is placed in an electric field, it experiences a force from higher to lower potential. On the other hand, a negative charge experiences a force from lower to higher potential.

The work done in moving a charge between two points in an electric field is independent of path followed between these two points, since the electric field is a conservative field.$\\$If a charged particle is accelerated through a potential difference of V volts, then the kinetic energy acquired by the particle is given by$\\$qV=$\\frac{1}{2}mv^2$$\\$⸫v=\sqrt{\frac{2qV}{m}$\\$Kinetic energy acquired by charged particles accelerated from rest through the same potential difference V is$\\$K.E α qat constant V$\\$

The ratio of the velocities acquired by two charged particles accelerated from rest through same potential difference are in the ratio$\\$$\frac{v₁}{v₂}=\sqrt{\frac{q₁}{q₂}×\frac{m₂}{m₁}}$$\\$i.e. v ∝$\sqrt{specific charge} \\$

Electric potential in few special cases:$\\$

The electric potential at a distance r from point charge q is given by$\\$V=\frac{1}{4πεₒ\frac{q}{r}

The electric potential at a point due to group of point charges $q_1, q_)2…......... q_n$ which are at distances $r_1, r_2…................r_n$ respectively is given by$\\$V=$V_1+V_2+…................+V_n \\$(Since electric potential is a scalar quantity, add directly)$\\$

The electric potential at any point due to dipole is$\\$V=$\frac{1}{4πεₒ}\frac{pcosθ}{r^2}$ ------> V∝$\frac{1}{r^2}$$\\$

where, p=q(2I)= electric dipole moment.$\\$

(i) The potential on the axial line of the dipole$\\$V=$\frac{1}{4πεₒ}\frac{p}{r^2}$$\\$(ii) The potential on the perpendicular bisector of dipole is zero but$\\$E=$\frac{1}{4πεₒ} \frac{p}{r^3} \\$

The electric field at a point is related to electric potential as -ve gradient of electric potential i.e.$\\$E=-$\frac{dV}{dr}$$\\$The -ve sign shows that electric field intensity E points in the direction of decreasing electric potential.$\\$

Do you know?$\\$Unit of E is N$C^{-1}$ or V$m^{-1} \\$It should be clear that electric potential is a scalar quantity, but electric potential gradient is a vector, as it is numerically equal to electric field intensity.

Electric potential due to charged conductor

For charged hollow spherical conductor$\\$(i) $V_{outside} = \frac{q}{4πε_0 r}$ (for r>R)$\\$=$\frac{σR^2}{ε_0 r}$$\\$(ii) $V_{surface} = V_{inside}$ &= $\frac{q}{4πε_0 R}$ ( for r<R or r=R)$\\$= $\frac{σR}{ε_0} \\$Where σ is the surface charge density$\\$(σ=$\frac{q}{A}=\frac{q}{4πR^2}$)$\\$

For non- conducting solid sphere$\\$(i) $V_{outside} = \frac{q}{4πε_0 r}$ ( for r>R)$\\$(ii) $V_{inside} = \frac{q}{4πε_0} (\frac{3R^2-r^2}{2R^3})$ ( for r<R)$\\$(iii) $V_{surface} =\frac{q}{4πε_0 R}$ ( for r=R)$\\$(iv) $V_{centre}= \frac{3}{2} \frac{q}{4πεₒR}$ (r=0)

Electric Potential Energy

Electric potential energy of a system of point charges is the total amount of work done in bringing the various charges to their respective positions from infinitely large mutual separations.

When two charges $q_1$ and $q_2$ are separated by a distance r, then potential energy of the system is U=$\frac{1}{4πε_0} \frac{q_1.q_2}{r}$

When three charges $q_1$, $q_2$ and $q_3$ are arranged at the three vertices of an equilateral triangle of side r, then P.E. of the system is given by

Similarly for square of side l,

The potential energy may be positive or negative depending on whether the work is done against the electric force or by the electric force during transportation of charge respectively.

Important Note:$\\$

To find potential energy at a point$\\$P.E(U)=q⋅$V_{net} \\$where q= charge which we are bringing from outside to inside.$\\$$V_{net}$ = Net potential at that point (the point at which we bring the charge.)

Equipotential Surface

The locus of all points which are at the same potential is known as equipotential surface.

No work is done to move a charge from one point to another on equipotential surface.

Near an isolated point charge the equipotential surface is a sphere.

The work done to move a unit positive charge around a charge q along a circle of radius r is zero.

The electric lines of force are always normal to equipotential surface, since $\vec{E}$ should not have a component along the equipotential surface.

The surface of a charged conductor is always an equipotential surface whatever may be its shape.

POINTS

The sure test of electrification is repulsion. (Attraction may occur between two oppositely charged bodies or a charged body and an uncharged one. That is why in attraction, one is not sure if both bodies are charged. However in repulsion, there are two bodies that have the same charge. That is why repulsion is the sure test of electrification.)

An electric charge at rest produces only electric field.

A thin stream of water from a tap is attracted by placing a rod near it. The rod is charged with static electricity.

Direction of electric lines of force is from +q to -q while direction of electric dipole is from -q to +q.

No work is done in carrying a charge from one point to another in an equipotential surface.

If a soap bubble is given a- ve charge, then its radius increases.

Electric Field Intensity ∝ No. of electric lines of force.

In a uniform electric field, an electric dipole experiences only torque but no space.$\\$
But torque may be zero if θ=0$\\$Since, τ=PEsin⁡$0^o$ = 0

But in non-uniform electric field, an electric dipole experiences both force and torque.

No point charge produces electric field at its own location. It produces electric field only outside it.

In case of a charged conductor:$\\$

i. Charge resides only on the outer surface of conductor.$\\$

ii. Electric field at any point inside the conductor is zero.$\\$

iii. Electric potential at any point inside the conductor is constant and equal to potential on the surface of the conductor, whatever be the shape and size of the conductor.$\\$

iv. Electric field at any point on the surface of charged conductor is directly proportional to the surface density of charge at that point, but electric potential does not depend upon the surface density of charge.$\\$

Due to point charge, E∝$\frac{1}{r^2}$ and V∝$\frac{1}{r}$

Due to electric dipole, E∝$\frac{1}{r^3}$ and V∝$\frac{1}{r^2}$

Due to quadrupole, E∝$\frac{1}{r^4}$

Due to an infinite uniform line of charge, E∝$\frac{1}{r}$ andV∝ $log_e$⁡r

Due to an infinite plane sheet of charge E∝$r^0$ and V∝r

On the surface of an irregularly shaped charged conductor, electric intensity may be different at different points but electric potential is same at every point.

For a dipole, $E_{axial} = 2E_{broad side}$ and $V_b$ = 0

Electric potential of earth is taken as zero because earth is a big conductor.

If an isolated conducting sphere is given a + ve charge, its mass decreases.

Electric field intensity due to a point charge $q_1$ at a distance $t_1$ and $t_2$ where $t_1$ is thickness of medium of dielectric constant $K_1$ and $t_2$ is thickness of medium of dielectric constant $K_2$ is$\\$E=$\frac{1}{4πε_0} \frac{q}{(t_1 \sqrt{K_1} + t_2(\sqrt{K_1})^2}$$\\$Electric potential at the same point is$\\$V=$\frac{1}{4πε_0} \frac{q}{(t_1 \sqrt{K_1} + t_2(\sqrt{K_1})}$$\\$

Two +ve charges are placed at points A and B. If one moves from point A to point B, the electric potential first decreases then increases.

If two conducting spheres of radii $r_1$ and $r_2$ are charged to the same surface charge density.$\\$The ratio of electric fields near their surface is 1: 1.$\\$

[Hint: $σ_1=σ_2$$\\$

∴ $\frac{Q_1}{4πr_1^2 }=\frac{Q_2}{4πr_2^2}$$\\$

" or " $\frac{Q_1}{4πε_or_1^2}$ = $\frac{Q_2}{4πε_or_2^2}$ ∴$E_1=E_2$ ]$\\$

The angle between the electric dipole moment and electric field due to it on the axial line is $0^o$, while on the broad side on position angle is $180^o$ i.e. $\vec{p}$ and $\vec{E}$ will be antiparallel to each other.

The electric potential produced by dipole at broad side position is zero.

If one penetrates a uniformly charged spherical cloud, electric field strength decreases directly as the distance from the centre.

When three equal like charges are placed at the vertices of an equilateral triangle, then the intensity at the intersection of the medians will be zero.

When two equal and unlike charges are placed at the two vertices of an equilateral triangle of side a, then the intensity of the electric field at the third vertex is given by$\\$E=$\frac{1}{4πε_o} \frac{q}{a^2}$

When two equal & like charges are placed at the two vertices of an equilateral triangle, then the intensity of the electric field at the third vertex is √3 E where$\\$E=$\frac{1}{4πε_o} \frac{q}{a^2}$

When four charges of q Coulomb each are placed at the four vertices of a square, then the intensity of the electric field at the intersection of the diagonals is zero.

Inside a hollow charged spherical conductor, the potential is constant and is equal to that of surface.

If two conducting spheres are separately charged and then brought in contact, the total charge on the two spheres is conserved.

In bringing an electron towards another electron, the electrostatic potential energy of the system increases.

A straight wire of length l and electric dipole moment p is bent to from a semicircle. The new dipole moment would be $\frac{2p}{π}$.

Electric flux ϕ is independent of the radius of the closed surface.

At the centre of a charged ring of radius R, E=0 and E is maximum at a distance $\frac{R}{√2}$ on the axis of the charged ring.

If electric potential energy is + ve, then force is repulsive. If it is -ve, then force is attractive and if force is zero, then system is equipotential.

Electrostatic energy density$\\$= $\frac{Electrostatic \space energy} {Volume }=\frac{1}{2} Kε_0 E^2$

For two equal charges of mass $m_1$ and $m_2$ to travel equal distances in a uniform electric field, if time taken is $t_1$ and $t_2$ respectively,$\\$$\frac{t_1^2}{m_1} =\frac{t_2^2}{m_2}$

Van de Graff generator is used to produce high DC potential (≈$10^6$ volts).

An electric charge$\\$

i. At rest produces only electric field$\\$ii. In uniform motion produces localized electric and magnetic fields.$\\$iii. Under acceleration/ deceleration produces electric and magnetic fields that are radiated.$\\$

When a charge is held stationary between horizontal plates having a p.d. of V volts$\\$qE=mg⇒q⋅$\frac{V}{d}$=mg

If two spheres of radii $R_1$ and $R_2$ carrying charges $Q_1$ and $Q_2$ are joined by a wire, then common potential$\\$V=$\frac{1}{4πε_0} \frac{Q_1+Q_2}{R_1+R_2}$

If two spherical conductors having radii $R_1$ and $R_2$ have same potential, then$\\$$\frac{σ_1}{σ_2} =\frac{R_2}{R_1}$$\\$

Two spheres of radii $R_1$ and $R_2$ respectively are charged and joined by a wire. Then,$\\$$\frac{E_1}{E_2} = \frac{R_2}{R_1}$ and $\frac{q_1}{q_2} = \frac{R_2}{R_1}$

The Gaussian surface for calculating the electric field due to a charge distribution is a symmetrical closed surface at every point of which electric field has a single fixed value.

Variation of electric field and potential is as follows:

Sharp edges or points are strictly avoided in electrical machines to avoid the loss of charges into the atmosphere.

The lighting conductors are made pointed upward in order to have a safe discharge of the electricity generated in the atmosphere.

When 8 drops of Hg each of 1V coalesces to form 1 drop, the potential of larger drop is 4V [Hint: V=$n^{2/3}$.V= $8^{2/3}$ $\space$ V=$(2)^{3×2/3}$ V=$2^2$V = 4V]

Charges are produced due to actual transfer of electrons.

Positive charge means the deficiency of electrons while negative charge means excess (gain) of electrons. So, during charge transfer there is change in mass too. Thus, mass of -ve-ly charged body is greater than that of +ve-ly charged body.