The area of a metallic plate is 10^{-2} m^2 . A charge of 10 \mu C is given to the plate. Determine the electric filed intensity at a point near to the plate.

Given A = 10^{-2} m^2 q = 10 \mu C = 10 \times 10^{-6} C Electric field intensity near plate E = \frac { \sigma}{ 2 \epsilon_0 } Where \sigma is surface charge density. \sigma = \frac {q}{A} \therefore \quad E = \frac {q}{ 2 \epsilon_0A} = \frac { 10 \times 10^{-6} }{ 2 \times 8.86 \times 10^{-12} \times 10^{-2} } = \frac {10^{-5} \times 10^{14} } {17.72} = 5.65 \times 10^7 V/m

Two point dipoles p\hat{k} and \dfrac{p}{2}\hat{k} are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).

The given point is on axis of \displaystyle \frac{\bar{p}}{2} dipole and at equatorial line of \bar{p} dipole so that field at given point is \displaystyle \left ( \bar{E_{1}}+ \bar{E_{2}} \right ) \displaystyle \bar{E_{1}}=\frac{2K(p/2)}{2^{3}}=\frac{Kp}{8}(+\hat{k}) \displaystyle \bar{E_{2}}=\frac{Kp}{1}(-\hat{k}) \displaystyle \bar{E_{1}}+\bar{E_{2}}=\frac{7}{8}Kp(-\hat{k})=-\frac{7p}{32\pi \varepsilon _{0}}\hat{k}

Electric Charges And Fields - Most Asked Questions

Q: A ball of charge q is placed in a hollow conducting uncharged sphere. After this, the sphere is connected with earth for a short time and the ball has not been brought into contact with the sphere. What charge will the sphere have after these operation? Where and how will this charge be distributed?

When a charged body is brought close to an uncharged body but they don't touch each other, then the surface of the uncharged body gets a charge of opposite sign and the bodies start attracting each other.This is known as charge induction.When the ball is inside the uncharged sphere, then the inner walls of the sphere will develop a charge of opposite type, whereas the outer walls will develop a charge of the same type. This is due to charge induction.Since the sphere is a good conductor, the charges will be equally distributed on both the sides.If the sphere is connected to earth, then the outer walls will lose all charge.If the ball doesn't touch the sphere, then the inner charges of the sphere will remain as it is.

Q: When a body is charged by induction, then the body

\bullet In electrostatic induction charge redistribute because of nearby charges. \bullet Charging by induction involves the transfer of charges from one part to the other part of the body (Inside body). \bullet Charge on an isolated system will always be conserved. \bullet Therefore there will be no loss or gain of charge. \bullet Hence option B is correct.

Q: How did Coulomb find the law of value of electric force between two point charges ?

Coulomb stated that the force (F) of attraction or repulsion between two points charges q_1 and q_2 separated by a distance r is proportional to the product of the magnitude of the two charges, and inversely proportional to the square of the distance of separation between them. i.e. F \propto \dfrac{q_1 q_2}{r^2}

Q: Two charges each of 100 micro coulomb are separated in a medium of relative permittivity 2 by a distance of 5 cm . The force between them is :

We know force between two charges in a medium separated by a distance 'r' F=\dfrac {(q_1)(q_2)}{4\pi \varepsilon r^2} where, \epsilon_r= \dfrac{\epsilon}{\epsilon_o} given, \epsilon_r=2 \implies \epsilon = 2\times \epsilon_o F=\dfrac {(100\times 10^{-6})(100\times 10^{-6})\times 9\times 10^9}{2\times (5\times 10^{-2})^2} F=\dfrac {90}{2\times 25\times 10^{-4}} F=\dfrac {90\times 100\times 10^2}{50} F=1\cdot 8\times 10^4 N

Q: Drawings I and II show two samples of electric field lines. Then :

The magnitude of electric field at a point can be determined by knowing how closely are the electric field lines packed at that particular point.In case 1: the field lines are equally spaced everywhere.In case 2: the field lines are spaced more closely in the right than in the left.

Q: Two electric field lines cannot cross each other. Also, they cannot form closed loops. Give reasons.

According to properties of electric field lines, The tangent of the field line given direction of electric field, there fore At a particular point electric field have one direction so field line never intersect.Also field lines originate from positive charge and end at a negative charge. Therefore it will not make close loop.

Q: Define electric flux. Write its SI unit.

Electric flux is defined as the measure of count of number of electric field lines crossing an area. Electric flux \phi = E A\cos\theta SI unit of electric flux is Nm^2/C

Q: Derive an expression for electric field due to an electric dipole at a point on its axial line.

Electric field due to an electric dipole at a point on its axial line: AB is an electric dipole of two point charges -q and +q separated by small distance 2d . P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.The electric fiedl at the point P due to +q placed at B is, { E }_{ 1 }=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { q }{ { \left( r-d \right) }^{ 2 } } (along BP )The electric field at the point P due to -q placed at A is, { E }_{ 2 }=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { q }{ { \left( r+d \right) }^{ 2 } } (along PA )Therefore, the magnitude of resultant electric field \left( E \right) acts in the direction of the vector with a greater, magnitude. The resultant electric field at P is E={ E }_{ 1 }+\left( -{ E }_{ 2 } \right) E=\left[ \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { q }{ { \left( r-d \right) }^{ 2 } } - \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { q }{ { \left( r+d \right) }^{ 2 } } \right] along BP E=\dfrac { q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 1 }{ { \left( r-d \right) }^{ 2 } } - \dfrac { 1 }{ { \left( r+d \right) }^{ 2 } } \right] along BP E=\dfrac { q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \dfrac { 4rd }{ { \left( { r }^{ 2 }-{ d }^{ 2 } \right) }^{ 2 } } \right] along BP If the point P is far away from the dipole, then d \ll r \therefore E=\dfrac { q }{ 4\pi { \varepsilon }_{ 0 } } - \dfrac { 4rd }{ { r }^{ 4 } } = \dfrac { q }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { 4d }{ { r }^{ 3 } } E=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { 2p }{ { r }^{ 3 } } along BP [ \because Electric dipole moment p=q\times 2d ] E acts in the direction of dipole moment.

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Electric Charges And Fields

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Charging by Induction

In this process, a charged object is brought near but not touched to a neutral conducting object. The presence of a charged object near a neutral conductor will induce (force) electrons within the conductor to move.
The movement of electrons leaves an imbalance of charge on opposite sides of the neutral conductor. While the overall object is neutral (i.e., has the same number of electrons as protons), there is an excess of positive charge on one side of the object and an excess of negative charge on the opposite side of the object.

A ball of charge

q

is placed in a hollow conducting uncharged sphere. After this, the sphere is connected with earth for a short time and the ball has not been brought into contact with the sphere. What charge will the sphere have after these operation? Where and how will this charge be distributed?

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When a body is charged by induction, then the body

becomes neutral

does not lose any charge

loses whole of the charge on it

loses part of the charge on it

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Vector form of coulomb's law

According to Coulomb's Law, electrostatic force on

q_{1}

due to

q_{2}

is given by:

F_{12} = \frac{1}{4 π ε _{o}} \frac{q _{1} q _{2}}{r _{12}^{2}} \overset{r_{21}}{^}

where

ϵ_{o} :

Permittivitty of the medium

q_{1}

and

q_{2}

: Magnitudes of Charges

r_{12} :

Distance between charges

\overset{r_{21}}{^} :

Unit vector directed from

q_{2}

q_{1}

How did Coulomb find the law of value of electric force between two point charges ?

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Two charges each of $100$ micro coulomb are separated in a medium of relative permittivity $2$ by a distance of $5 c m$ . The force between them is :

0.36 \times 1 0^{5} N

3.6 \times 1 0^{5} N

1.8 \times 1 0^{4} d y n e

1.8 \times 1 0^{4} N

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Properties of electric field lines

The field lines follow some important general properties:
(i) Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity.
(ii) In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
(iii) Two field lines can never cross each other. (If they did,the field at the point of intersection will not have a unique direction, which is absurd.)
(iv) Electrostatic field lines do not form any closed loops.

Drawings I and II show two samples of electric field lines. Then :

the electric fields in both I and II are produced by negative charge located somewhere on the left and positive charges located somewhere on the right

in both I and II the electric field is the same everywhere

in both cases the field becomes stronger on moving from left to right

the electric field in I is the same everywhere, but in II the electric field becomes stronger on moving from left to right

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Two electric field lines cannot cross each other. Also, they cannot form closed loops. Give reasons.

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Electric flux

Electric flux is the measure of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a normally perpendicular surface.

Define electric flux. Write its SI unit.

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The area of a metallic plate is

1 0^{- 2} m^{2}

. A charge of

10 μ C

is given to the plate. Determine the electric filed intensity at a point near to the plate.

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Electric field due to a dipole

E = \frac{1}{4 π ϵ _{0}} \frac{p}{r ^{3}} 3 c o s^{2} θ + 1

where

θ

is the angle between the distance vector and dipole.

Derive an expression for electric field due to an electric dipole at a point on its axial line.

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Two point dipoles

p \hat{k}

and

\frac{p}{2} \hat{k}

are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).

\frac{9 p}{3 2 π ε _{0}} \hat{k}

\frac{- 7 p}{3 2 π ε _{0}} \hat{k}

\frac{7 p}{3 2 π ε _{0}} \hat{k}

none of these

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Electric field due to an infinite linear charge distribution

We have the charge

q_{e n c}

enclosed by the
cylinder. Because the linear charge density (charge per unit length, remember) is
uniform, the enclosed charge is

λ

h.Thus, Gauss law

ϵ_{o} Φ = q_{e n c l o s e d}

ϵ_{o} E (2 π r h) = λ h

E = \frac{λ}{2 π ϵ _{0} r}

where

λ

is the linear charge density and r is the distance at which the electric field is to be calculated.

Use Gauss's theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density

σ

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The region between two concentric spheres of radii 'a' and 'b', respectively(see figure), has volume charge density

ρ = \frac{A}{r}

, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :

\frac{Q}{2 π a ^{2}}

\frac{Q}{2 π ( b ^{2} - a ^{2} )}

\frac{2 Q}{π ( a ^{2} - b ^{2} )}

\frac{2 Q}{π a ^{2}}

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