How do the density and direction of electric field lines around a charge relate to the magnitude and sign of that charge?

The density of electric field lines represents the magnitude of the electric field. Closer lines indicate a stronger field. The direction of the field lines indicates the direction of the force on a positive test charge. Electric field lines radiate outward from positive charges and converge inward toward negative charges.

Why do electric fields behave differently in and around conductors compared to insulators?

In conductors, free charges (electrons) move in response to an electric field, rearranging until the field inside is zero. The field lines end perpendicularly on the surface. In insulators, charges can't move, so the electric field can exist within the material.

How does a conductive enclosure shield its interior from external electric fields, and what are some practical applications of this phenomenon?

Free charges in a conductor rearrange to cancel external electric fields, with field lines terminating on the surface to block penetration. This principle is used in shielding electronic devices and Faraday cages.

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Electric Field

The electric field is a fundamental idea in physics that helps us understand how electric charges affect each other. It explains the forces of attraction or repulsion between charges through the space around them. This concept was first introduced by Michael Faraday and later developed more fully by James Clerk Maxwell. The electric field is crucial in electromagnetism, as it shows how electric and magnetic fields interact and how they are influenced by charges.

1.0Electric Field Intensity

The space around a charge or charge distribution, in which another charge experiences an electric force, is called an electric field.
It is defined as the net force experienced by a unit positive test charge.

$E = \frac{F}{q _{0}}$

SI Unit N/C or V/m

Dimensional formula: $[M^{1} L^{1} T^{- 3} A^{- 1}]$

Test Charge : It is a charge of very small magnitude which does not produce a significant electric field.

$E = q_{0} \to 0 lim \frac{F}{q _{0}}$

Force on a point charge is in the same direction as that of electric field on positive charge and in opposite direction as that of electric field on a negative charge.

Electric Field due to a Point Charge

Electric field in vacuum $E = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}$

Electric field in Medium $E_{m} = \frac{1}{4 π ε} \frac{Q}{r ^{2}} = \frac{E}{K}$

Note: - K is dielectric constant of medium also known as relative permittivity of medium.

$(K \geq 1) or (E_{m} \leq E)$

2.0Electric Field Owed to Positive and Negative Charge

3.0Electric Field due to System of Charges

Electric field follows superposition principle,When two or more than two charges are present in space then the electric field at a particular point will be identical to the vector sum of electric fields due to individual charges.

$E_{net} = E_{1} + E_{2} + E_{3} + \dots + E_{n}$

Important Note: When identical charges are placed on the corners of a regular polygon (symmetric arrangement) then the resultant field at the centre of the polygon is always zero.

4.0Graphical Problems Electric Field

Electric field v/s distance

For point charge $E = \frac{k Q}{r ^{2}} or E \propto \frac{1}{r ^{2}}$

(1) For positive charge :

Graphical problems Electric Field (For positive charge)

(2) For negative charge

Graphical problems Electric Field (For negative charge)

(3) For Positive and Negative Charge (combined analysis)

Graphical problems Electric Field (For positive and negative charge (combined analysis))

(4) Graph for pair of positive charges

Graphical problems Electric Field (Graph for pair of positive charges )

(5) Graph for pair of negative charges

Graphical problems Electric Field ( Graph for pair of negative charges)

(6) Graph for pair of positive and negative charges

Graphical problems Electric Field ( Graph for pair of positive and negative charges)

Here M is the midpoint of the line joining the two charges at x distance from any charge.Here electric field will be E=2kQx2 (E = \frac{2 kQ}{x^2}) (rightward minimum electric field)

5.0Continuous Charged Distribution

Linear Charge distribution	Surface Charge distribution	Volume Charge distribution

Used for linear objects (1-D) such as wire, thin rod, ring etc.	Used for flat objects (2-D) such as plate, disc, etc.	Used for 3-D objects such as sphere, cube, cylinder etc.
Linear charge density is, $λ = (\frac{Q}{L}) C / m$	Surface charge density, $σ = (\frac{Q}{A}) C / m^{2}$	Volume charge density, $ρ = (\frac{Q}{V}) C / m^{3}$

6.0Electric field intensity due to charged wire

(1) For infinite wire, (both ends goes to infinite)

$θ_{1} \to 9 0^{\circ}, θ_{2} \to 9 0^{\circ}$

$E_{x} = E_{⊥} = \frac{kλ}{r} (sin 9 0^{\circ} + sin 9 0^{\circ}) = \frac{2 kλ}{r}$

$E_{y} = E_{∥} = \frac{kλ}{r} (cos 9 0^{\circ} + cos 9 0^{\circ}) = 0$

$E_{x} = E_{⊥} = \frac{2 kλ}{r} (perpendicular to wire)$

(2) For semi – Infinite wire

$E_{x} = E_{l} = \frac{kλ}{r} (sin 0^{\circ} + sin 9 0^{\circ}) = \frac{kλ}{r}$

$E_{y} = E_{l} = \frac{kλ}{r} (cos 0^{\circ} - cos 9 0^{\circ}) = \frac{kλ}{r}$

$E_{r} = \frac{2 kλ}{r}$

(3) Electric field due to finite wire at symmetric point

$E_{x} = E_{l} = \frac{kλ}{r} (sin θ + sin θ) = \frac{2 kλ}{r} sin θ$

$E_{y} = E_{l} = \frac{kλ}{r} (cos θ - cos θ) = 0$

7.0Electric Field due to a Uniformly Charged Arc

$E = \frac{2 kλ}{r} sin (\frac{θ}{2})$ is linear charge density

8.0Electric Field at Centre of Evenly Charged Ring

$E = \frac{2 kλ}{r} sin (\frac{θ}{2}), θ \to angle of arc$

$E = \frac{2 kλ}{r} sin (\frac{36 0 ^{\circ}}{2}) = 0$

9.0Electric Field due to a Evenly Charged Ring at its Axis

$E = \frac{- k Q x}{( R ^{2} + x ^{2} ) ^{\frac{3}{2}}}$

(1) Electric field on the axis for small values of x

$E = \frac{k Q x}{R ^{3}}$

(2) Electric field at the centre of ring,E will be zero at x=0

(3) Electric field at the axis for larger values of x,

$E = \frac{k Q}{x ^{2}} (Ring behaves like a point charge)$

(4) Maximum value of electric field, $(x = \frac{R}{2})$

So, $E_{ma x} = E$ $at x = \frac{R}{2}, we have E_{m a x} = \frac{- 2 k Q}{3 3 R ^{2}}$

Variation of Electric Field for Ring

10.0Electric Field on the Axis of an Evenly Charged Disc

$E_{p} = \frac{σ}{2 ε _{0}} [1 - cos θ]$

$E_{p} = \frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]$

11.0Electric Dipole and Electric Field Owing to a Dipole

A system of two equal and opposite charges placed at very small separation is known as electric dipole.

Dipole Moment:

The dipole moment of a dipole is equal to the product of magnitude of either charge and separation between the charges.

$p = q l$

It is a vector quantity whose direction is from (-q) to (+q)

SI unit → C-m,

Practical unit → debye $(3.33 \times 1 0^{- 30} C - m)$

Electric Field Due to a Dipole

(1) At Axial / End on position

$E_{Axial} = \frac{2 k p}{r ^{3}}$

(2) At Equator / Broad Side on position

$E_{Equi} = \frac{k p}{r ^{3}}$

(3) At General Position

$E = \frac{k p}{r ^{3}} 3 cos^{2} θ + 1$

$tan α = \frac{1}{2} tan θ$

12.0Electric Field Lines

These are imaginary lines used to represent electric fields pictorially.These are such a smooth continuous curve that tangent drawn at any point gives direction of the field at that point.

Properties of Electric Field Lines:

(1) They originate from (+) charge and terminate at (-)charge.

(2)It gives an idea about the magnitude of charge.

$∣ q ∣ (\propto) number of field lines$

(3) It gives an idea about the strength of the electric field.

Density of electric field lines

$(EF L) \propto strength of intensity of electric field.$ n

If electric field lines are denser electric field intensity is high.
If electric field lines are rarer electric field intensity is low.

(4) Tangent drawn at any point of the electric field line,gives the direction of force on a charge particle placed at that given point.

Note: It gives the direction of force not the direction of motion.

(5) Two electric field lines can never intersect each other.

Reason: If they do so, then at the point of intersection, there will be two tangents at the same point representing two different directions of electric field at the point which is not possible.

(6) Electrostatic field lines can never form any closed loop, but induced electric field lines can form a closed loop.

(7) Electric field lines due to a pair of like nature of charges.

(8) Electric field lines due to a pair of equal and opposite charges.

13.0Solved Examples

1. Charge Q is uniformly spread over a ring of radius R. If a small portion of length d is removed from the ring, then find an electric field at the centre of the ring.

Solution:

If dq is charge of small element then,

$E = \frac{k d q}{R ^{2}} = \frac{kλ d l}{R ^{2}} = \frac{k d l}{R ^{2}} (\frac{Q}{2 π R})$

2. Find electric field on the axis of a uniformly charged ring (Q,R) at a distance $x = 3 R$

Solution:

$E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{\frac{3}{2}}}$

$A t (x = 3 R)$

$E = \frac{k Q x}{( R ^{2} + ( 3 R ) ^{2} ) ^{\frac{3}{2}}} 3 R \Rightarrow E = \frac{3 k Q}{8 R ^{2}}$

Join ALLEN!

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Class

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Electric Field

The electric field is a fundamental idea in physics that helps us understand how electric charges affect each other. It explains the forces of attraction or repulsion between charges through the space around them. This concept was first introduced by Michael Faraday and later developed more fully by James Clerk Maxwell. The electric field is crucial in electromagnetism, as it shows how electric and magnetic fields interact and how they are influenced by charges.

1.0Electric Field Intensity

The space around a charge or charge distribution, in which another charge experiences an electric force, is called an electric field.
It is defined as the net force experienced by a unit positive test charge.

$E = \frac{F}{q _{0}}$

SI Unit N/C or V/m

Dimensional formula: $[M^{1} L^{1} T^{- 3} A^{- 1}]$

Test Charge : It is a charge of very small magnitude which does not produce a significant electric field.

$E = q_{0} \to 0 lim \frac{F}{q _{0}}$

Force on a point charge is in the same direction as that of electric field on positive charge and in opposite direction as that of electric field on a negative charge.

Electric Field due to a Point Charge

Electric field in vacuum $E = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}}$

Electric field in Medium $E_{m} = \frac{1}{4 π ε} \frac{Q}{r ^{2}} = \frac{E}{K}$

Note: - K is dielectric constant of medium also known as relative permittivity of medium.

$(K \geq 1) or (E_{m} \leq E)$

2.0Electric Field Owed to Positive and Negative Charge

3.0Electric Field due to System of Charges

Electric field follows superposition principle,When two or more than two charges are present in space then the electric field at a particular point will be identical to the vector sum of electric fields due to individual charges.

$E_{net} = E_{1} + E_{2} + E_{3} + \dots + E_{n}$

Important Note: When identical charges are placed on the corners of a regular polygon (symmetric arrangement) then the resultant field at the centre of the polygon is always zero.

4.0Graphical Problems Electric Field

Electric field v/s distance

For point charge $E = \frac{k Q}{r ^{2}} or E \propto \frac{1}{r ^{2}}$

(1) For positive charge :

(2) For negative charge

(3) For Positive and Negative Charge (combined analysis)

(4) Graph for pair of positive charges

(5) Graph for pair of negative charges

(6) Graph for pair of positive and negative charges

Here M is the midpoint of the line joining the two charges at x distance from any charge.Here electric field will be E=2kQx2 (E = \frac{2 kQ}{x^2}) (rightward minimum electric field)

5.0Continuous Charged Distribution

Linear Charge distribution	Surface Charge distribution	Volume Charge distribution

Used for linear objects (1-D) such as wire, thin rod, ring etc.	Used for flat objects (2-D) such as plate, disc, etc.	Used for 3-D objects such as sphere, cube, cylinder etc.
Linear charge density is, $λ = (\frac{Q}{L}) C / m$	Surface charge density, $σ = (\frac{Q}{A}) C / m^{2}$	Volume charge density, $ρ = (\frac{Q}{V}) C / m^{3}$

6.0Electric field intensity due to charged wire

(1) For infinite wire, (both ends goes to infinite)

$θ_{1} \to 9 0^{\circ}, θ_{2} \to 9 0^{\circ}$

$E_{x} = E_{⊥} = \frac{kλ}{r} (sin 9 0^{\circ} + sin 9 0^{\circ}) = \frac{2 kλ}{r}$

$E_{y} = E_{∥} = \frac{kλ}{r} (cos 9 0^{\circ} + cos 9 0^{\circ}) = 0$

$E_{x} = E_{⊥} = \frac{2 kλ}{r} (perpendicular to wire)$

(2) For semi – Infinite wire

$E_{x} = E_{l} = \frac{kλ}{r} (sin 0^{\circ} + sin 9 0^{\circ}) = \frac{kλ}{r}$

$E_{y} = E_{l} = \frac{kλ}{r} (cos 0^{\circ} - cos 9 0^{\circ}) = \frac{kλ}{r}$

$E_{r} = \frac{2 kλ}{r}$

(3) Electric field due to finite wire at symmetric point

$E_{x} = E_{l} = \frac{kλ}{r} (sin θ + sin θ) = \frac{2 kλ}{r} sin θ$

$E_{y} = E_{l} = \frac{kλ}{r} (cos θ - cos θ) = 0$

7.0Electric Field due to a Uniformly Charged Arc

$E = \frac{2 kλ}{r} sin (\frac{θ}{2})$ is linear charge density

8.0Electric Field at Centre of Evenly Charged Ring

$E = \frac{2 kλ}{r} sin (\frac{θ}{2}), θ \to angle of arc$

$E = \frac{2 kλ}{r} sin (\frac{36 0 ^{\circ}}{2}) = 0$

9.0Electric Field due to a Evenly Charged Ring at its Axis

$E = \frac{- k Q x}{( R ^{2} + x ^{2} ) ^{\frac{3}{2}}}$

(1) Electric field on the axis for small values of x

$E = \frac{k Q x}{R ^{3}}$

(2) Electric field at the centre of ring,E will be zero at x=0

(3) Electric field at the axis for larger values of x,

$E = \frac{k Q}{x ^{2}} (Ring behaves like a point charge)$

(4) Maximum value of electric field, $(x = \frac{R}{2})$

So, $E_{ma x} = E$ $at x = \frac{R}{2}, we have E_{m a x} = \frac{- 2 k Q}{3 3 R ^{2}}$

Variation of Electric Field for Ring

10.0Electric Field on the Axis of an Evenly Charged Disc

$E_{p} = \frac{σ}{2 ε _{0}} [1 - cos θ]$

$E_{p} = \frac{σ}{2 ε _{0}} [1 - \frac{x}{R ^{2} + x ^{2}}]$

11.0Electric Dipole and Electric Field Owing to a Dipole

A system of two equal and opposite charges placed at very small separation is known as electric dipole.

Dipole Moment:

The dipole moment of a dipole is equal to the product of magnitude of either charge and separation between the charges.

$p = q l$

It is a vector quantity whose direction is from (-q) to (+q)

SI unit → C-m,

Practical unit → debye $(3.33 \times 1 0^{- 30} C - m)$

Electric Field Due to a Dipole

(1) At Axial / End on position

$E_{Axial} = \frac{2 k p}{r ^{3}}$

(2) At Equator / Broad Side on position

$E_{Equi} = \frac{k p}{r ^{3}}$

(3) At General Position

$E = \frac{k p}{r ^{3}} 3 cos^{2} θ + 1$

$tan α = \frac{1}{2} tan θ$

12.0Electric Field Lines

These are imaginary lines used to represent electric fields pictorially.These are such a smooth continuous curve that tangent drawn at any point gives direction of the field at that point.

Properties of Electric Field Lines:

(1) They originate from (+) charge and terminate at (-)charge.

(2)It gives an idea about the magnitude of charge.

$∣ q ∣ (\propto) number of field lines$

(3) It gives an idea about the strength of the electric field.

Density of electric field lines

$(EF L) \propto strength of intensity of electric field.$ n

If electric field lines are denser electric field intensity is high.
If electric field lines are rarer electric field intensity is low.

(4) Tangent drawn at any point of the electric field line,gives the direction of force on a charge particle placed at that given point.

Note: It gives the direction of force not the direction of motion.

(5) Two electric field lines can never intersect each other.

Reason: If they do so, then at the point of intersection, there will be two tangents at the same point representing two different directions of electric field at the point which is not possible.

(6) Electrostatic field lines can never form any closed loop, but induced electric field lines can form a closed loop.

(7) Electric field lines due to a pair of like nature of charges.

(8) Electric field lines due to a pair of equal and opposite charges.

13.0Solved Examples

1. Charge Q is uniformly spread over a ring of radius R. If a small portion of length d is removed from the ring, then find an electric field at the centre of the ring.

Solution:

If dq is charge of small element then,

$E = \frac{k d q}{R ^{2}} = \frac{kλ d l}{R ^{2}} = \frac{k d l}{R ^{2}} (\frac{Q}{2 π R})$

2. Find electric field on the axis of a uniformly charged ring (Q,R) at a distance $x = 3 R$

Solution:

$E = \frac{k Q x}{( R ^{2} + x ^{2} ) ^{\frac{3}{2}}}$

$A t (x = 3 R)$

$E = \frac{k Q x}{( R ^{2} + ( 3 R ) ^{2} ) ^{\frac{3}{2}}} 3 R \Rightarrow E = \frac{3 k Q}{8 R ^{2}}$

Frequently Asked Questions

How do the density and direction of electric field lines around a charge relate to the magnitude and sign of that charge?

Why do electric fields behave differently in and around conductors compared to insulators?

How does a conductive enclosure shield its interior from external electric fields, and what are some practical applications of this phenomenon?

Frequently Asked Questions

How do the density and direction of electric field lines around a charge relate to the magnitude and sign of that charge?

Why do electric fields behave differently in and around conductors compared to insulators?

How does a conductive enclosure shield its interior from external electric fields, and what are some practical applications of this phenomenon?

Join ALLEN!

Electric Field

1.0Electric Field Intensity

Electric Field due to a Point Charge

2.0Electric Field Owed to Positive and Negative Charge

3.0Electric Field due to System of Charges

4.0Graphical Problems Electric Field

5.0Continuous Charged Distribution

6.0Electric field intensity due to charged wire

7.0Electric Field due to a Uniformly Charged Arc

8.0Electric Field at Centre of Evenly Charged Ring

9.0Electric Field due to a Evenly Charged Ring at its Axis

10.0Electric Field on the Axis of an Evenly Charged Disc

11.0Electric Dipole and Electric Field Owing to a Dipole

12.0Electric Field Lines

13.0Solved Examples

Table of Contents

Join ALLEN!

Electric Field

1.0Electric Field Intensity

Electric Field due to a Point Charge

2.0Electric Field Owed to Positive and Negative Charge

3.0Electric Field due to System of Charges

4.0Graphical Problems Electric Field

5.0Continuous Charged Distribution

6.0Electric field intensity due to charged wire

7.0Electric Field due to a Uniformly Charged Arc

8.0Electric Field at Centre of Evenly Charged Ring

9.0Electric Field due to a Evenly Charged Ring at its Axis

10.0Electric Field on the Axis of an Evenly Charged Disc

11.0Electric Dipole and Electric Field Owing to a Dipole

12.0Electric Field Lines

13.0Solved Examples

Table of Contents