Plane Mirror

Image formed is as far behind the mirror as the object is in front of it, with the lateral inversion of image. i.e. right hand side of object is left hand side of image and vice versa.

Convex Mirror

Image formed is always virtual, erect and diminished.

Concave Mirror

Depending upon the position of the object, size of the image formed can be equal to, larger than or smaller than the size of object. Also the nature of image formed may be real or virtual.

Plane Mirror

Image formed is as far behind the mirror as the object is in front of it, with the lateral inversion of image. i.e. right hand side of object is left hand side of image and vice versa.

Convex Mirror

Image formed is always virtual, erect and diminished.

Concave Mirror

Depending upon the position of the object, size of the image formed can be equal to, larger than or smaller than the size of object. Also the nature of image formed may be real or virtual.

The distance between the principal focus of a mirror and the pole of mirror is called focal length of the mirror.

In case of Concave mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as negative.

In case of Convex mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as positive.

The radius of Curvature of plane mirror is infinite, hence the focal length of plane mirror will also be infinite.

The distance between the principal focus of a mirror and the pole of mirror is called focal length of the mirror.

In case of Concave mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as negative.

In case of Convex mirror distance between the pole ‘O’ and principal focus ‘F’ of mirror is focal length and is always taken as positive.

The radius of Curvature of plane mirror is infinite, hence the focal length of plane mirror will also be infinite.

Solution: For a concave mirror u becomes negative, $$\therefore u = -7.5 cm $$ (i) When real image is formed v is also negative $$ \therefore v = -30 cm $$

By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ -30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { -30-7.5 }{ \left( -7.5 \right) \times \left( -30 \right) } = \frac { -37.5 }{ 225 } $$ $$ \therefore f = \frac { 225 }{ -37.5 } = -6 cm $$ (ii) When virtual image is formed v is positive $$ \therefore v = 30 cm $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ 30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { 30-7.5 }{ \left( -7.5 \right) \times \left( 30 \right) } $$ $$= \frac { 22.5 }{ -225 } $$ $$ \therefore f = \frac { -225 }{ 22.5 } = -10 cm $$ Hence the focal length of mirror when real image is formed is -6 cm and when virtual image is formed at 30 cm, focal length of mirror will be -10 cm.

Solution: For a concave mirror u becomes negative, $$\therefore u = -7.5 cm $$ (i) When real image is formed v is also negative $$ \therefore v = -30 cm $$

By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ -30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { -30-7.5 }{ \left( -7.5 \right) \times \left( -30 \right) } = \frac { -37.5 }{ 225 } $$ $$ \therefore f = \frac { 225 }{ -37.5 } = -6 cm $$ (ii) When virtual image is formed v is positive $$ \therefore v = 30 cm $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ -7.5 } +\frac { 1 }{ 30 } =\frac { 1 }{ f } $$ $$ \therefore \frac { 1 }{ f } = \frac { 30-7.5 }{ \left( -7.5 \right) \times \left( 30 \right) } $$ $$= \frac { 22.5 }{ -225 } $$ $$ \therefore f = \frac { -225 }{ 22.5 } = -10 cm $$ Hence the focal length of mirror when real image is formed is -6 cm and when virtual image is formed at 30 cm, focal length of mirror will be -10 cm.

Solution: As we have a concave mirror focal length is negative, $$ \therefore f = -50 cm $$ and $$ Magnification (m) = \frac { v }{ u } = 2 $$ $$ \therefore v = 2u $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ u } +\frac { 1 }{ 2u } =\frac { 1 }{ -50 } \frac { 2+1 }{ 2u } $$ $$=\frac { -1 }{ 50 } \frac { 3 }{ 2u } $$ $$ =\frac { -1 }{ 50 } $$ $$ \therefore 2u = -\left( 3 \times 50 \right) $$ or $$ u = \frac { -150 }{ 2 } = -75 cm $$

Solution: As we have a concave mirror focal length is negative, $$ \therefore f = -50 cm $$ and $$ Magnification (m) = \frac { v }{ u } = 2 $$ $$ \therefore v = 2u $$ By using mirror formula, $$ \frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } $$ Lets substitute the values, $$ \frac { 1 }{ u } +\frac { 1 }{ 2u } =\frac { 1 }{ -50 } \frac { 2+1 }{ 2u } $$ $$=\frac { -1 }{ 50 } \frac { 3 }{ 2u } $$ $$ =\frac { -1 }{ 50 } $$ $$ \therefore 2u = -\left( 3 \times 50 \right) $$ or $$ u = \frac { -150 }{ 2 } = -75 cm $$

(i) The pole of mirror is taken as origin.

(ii) All distances are measured from the pole ‘O’ of the mirror.

(iii) Rays of light are assumed to be coming from left side and enter the right side.

(iv) Distances measured in the direction of incident ray are taken as positive.

(v) All distances measured in direction opposite to incident ray are taken as negative.

(vi) Height measured above the principal axis are taken as positive and below as negative.

(vii) For concave mirrors, radius of curvature (R) and focal length (f) are taken as negative.

(viii) For convex mirrors, radius of curvature (R) and focal length (f) are taken as positive.

(ix) For real image magnification is negative and for virtual image magnification is positive.

(i) The pole of mirror is taken as origin.

(ii) All distances are measured from the pole ‘O’ of the mirror.

(iii) Rays of light are assumed to be coming from left side and enter the right side.

(iv) Distances measured in the direction of incident ray are taken as positive.

(v) All distances measured in direction opposite to incident ray are taken as negative.

(vi) Height measured above the principal axis are taken as positive and below as negative.

(vii) For concave mirrors, radius of curvature (R) and focal length (f) are taken as negative.

(viii) For convex mirrors, radius of curvature (R) and focal length (f) are taken as positive.

(ix) For real image magnification is negative and for virtual image magnification is positive.

If two charges \({ q }_{ 1 }\) and \({ q }_{ 2 }\) are separated by distance \(r\) in vacuum then electrostatic force between them will be given by the Coulomb law as below:

$${ F }_{ v } = \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(1)$$ Now if these charges are placed in a medium of absolute permittivity \(\epsilon\) , then electrostatic force between these charges will be $${ F }_{ m } = \frac { 1 }{ 4\pi \epsilon } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(2)$$ We know that $$\epsilon = { \epsilon }_{ 0 } { \epsilon }_{ r } =K{ \epsilon }_{ 0 }$$ where \(K\) is dielectric constant or Relative permittivity or specific inductive capacity. or $$K = \frac { \epsilon }{ { \epsilon }_{ 0 } }$$ Now, eqn. (2) becomes, $${ F }_{ m } = \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(3)$$ Dividing eqn. (1) by eqn. (3), $$\frac { { F }_{ v } }{ { F }_{ m } } = \frac { \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) }{ \left( \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) } = K$$ Hence, dielectric constant (or relative permittivity) of a medium is defined as the ratio of the electrostatic force between two point charges when placed a certain distance apart in vacuum to the electrostatic force between the same charges when placed same distance apart in a medium.

This means a dielectric reduces the force between two charges.

If two charges \({ q }_{ 1 }\) and \({ q }_{ 2 }\) are separated by distance \(r\) in vacuum then electrostatic force between them will be given by the Coulomb law as below:

$${ F }_{ v } = \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(1)$$ Now if these charges are placed in a medium of absolute permittivity \(\epsilon\) , then electrostatic force between these charges will be $${ F }_{ m } = \frac { 1 }{ 4\pi \epsilon } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(2)$$ We know that $$\epsilon = { \epsilon }_{ 0 } { \epsilon }_{ r } =K{ \epsilon }_{ 0 }$$ where \(K\) is dielectric constant or Relative permittivity or specific inductive capacity. or $$K = \frac { \epsilon }{ { \epsilon }_{ 0 } }$$ Now, eqn. (2) becomes, $${ F }_{ m } = \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(3)$$ Dividing eqn. (1) by eqn. (3), $$\frac { { F }_{ v } }{ { F }_{ m } } = \frac { \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) }{ \left( \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) } = K$$ Hence, dielectric constant (or relative permittivity) of a medium is defined as the ratio of the electrostatic force between two point charges when placed a certain distance apart in vacuum to the electrostatic force between the same charges when placed same distance apart in a medium.

This means a dielectric reduces the force between two charges.

Permittivity is the property of a particular medium which affects the magnitude of the force existing between two point charges.

We know that the greater the value of the permittivity of the medium placed between the two charged bodies the lesser the value of force existing between them.

Absolute permittivity \({ \epsilon }_{ 0 }\)

Relative permittivity \({ \epsilon }_{ r }\)

The absolute permittivity of air or vacuum is minimum and its value is \(8.854 \times { 10 }^{ -12 }\) F/m (farad/metre) whereas the value of absolute or (actual) permittivity \( \epsilon \) of all other insulating medium is more than \({ \epsilon }_{ 0 }\).

The ratio of these two permittivities i.e. absolute permittivity (\( \epsilon \)) of the insulating medium to the absolute permittivity (\({ \epsilon }_{ 0 }\)) of the air or vacuum is known as relative permittivity of that medium and is denoted by \({ \epsilon }_{ r }\). i.e. $$\quad { \epsilon }_{ r } = \frac { \epsilon }{ { \epsilon }_{ 0 } } $$

Permittivity is the property of a particular medium which affects the magnitude of the force existing between two point charges.

We know that the greater the value of the permittivity of the medium placed between the two charged bodies the lesser the value of force existing between them.

Absolute permittivity \({ \epsilon }_{ 0 }\)

Relative permittivity \({ \epsilon }_{ r }\)

The absolute permittivity of air or vacuum is minimum and its value is \(8.854 \times { 10 }^{ -12 }\) F/m (farad/metre) whereas the value of absolute or (actual) permittivity \( \epsilon \) of all other insulating medium is more than \({ \epsilon }_{ 0 }\).

The ratio of these two permittivities i.e. absolute permittivity (\( \epsilon \)) of the insulating medium to the absolute permittivity (\({ \epsilon }_{ 0 }\)) of the air or vacuum is known as relative permittivity of that medium and is denoted by \({ \epsilon }_{ r }\). i.e. $$\quad { \epsilon }_{ r } = \frac { \epsilon }{ { \epsilon }_{ 0 } } $$

Coulomb performed a number of experiments to see the effect of placing two small charges near each other. On his experimental researches, he established a law which is known as Coulombs Law of Electrostatics.

According to Coulombs Law :The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges \({ q }_{ 1 } \) and \({ q }_{ 2 }\) have distance d between them, then force F between the charges can be mathematically expressed as: $$ F \propto { q }_{ 1 }{ q }_{ 2 }\qquad ...(1)$$ $$ F \propto \frac { 1 }{ { d }^{ 2 } } \qquad ...(2)$$ Combining (1) and (2), $$ F \propto \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } or F = K\frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \qquad ...(3) $$ Where, \(K\) is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used.

In SI units force is measured in newton, charge in coulomb, distance in metre and the value of \(K\) is given as, $$ K = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \qquad ...(4) (in SI system)$$ where, \({ \epsilon }_{ 0 }\) = Absolute permittivity of vacuum or permittivity of free space = \(8.854 \times { 10 }^{ -12 }\) farad/metre
\( { \epsilon }_{ r }\) = Relative permittivity of the medium w.r.t. vacuum in which the charges are placed (e.g. for air \({ \epsilon }_{ r }\) = 1)

Putting the value of \(K\) in equation (3), we get, $$ \boxed { F = \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } \qquad ... (5)$$ Unit Charge or one coulomb charge in SI system can be defined as the amount of charge which when placed at a distance of one metre from an equal and similar charge in air, is repelled with a force of \(9 \times { 10 }^{ 9 } \)newton from it.

As per this defination, $$ { q }_{ 1 } = { q }_{ 2 } = q, $$ $$F = 9 \times { 10 }^{ 9 } newton,$$ $$ { \epsilon }_{ 0 } = 8.854 \times { 10 }^{ -12 } farad/metre,$$ \({ \epsilon }_{ r } = 1\) (Assuming two charges to be placed in air)
d = 1m

Putting these values in the above relation, we get $$ 9 \times { 10 }^{ 9 } = \frac { q \times q }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 \times { 1 }^{ 2 } } $$ $$ 9 \times { 10 }^{ 9 } = \frac { { q }^{ 2 } }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) } $$ $$ { q }^{ 2 } = \left( 9 \times { 10 }^{ 9 } \right) \times 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) = 1.00086 $$ $$ \therefore q = \pm 1.00043 \ coulomb $$ Hence, unit charge.

Determine the force between two charges, each of one coulomb when they are separated at one metre distance in air.

Solution:We have, $${ q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = 1m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \frac { 1 \times 1 }{ { \left( 1 \right) }^{ 2 } } $$ $$ F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } $$ $$ F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } $$ $$ F = 8.992 \times { 10 }^{ 9 } N $$

An electron and a proton are at a distance of 10^-9 m from each other in a free space. Compute the force between them.

Solution:We have, The charge on an electron, \({ q }_{ 1 } = -1.6 \times { 10 }^{ -19 } \) C
and the charge on a proton, \({ q }_{ 2 } = +1.6 \times { 10 }^{ -19 }\) C. $$ { q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = \({ 10 }^{ -9 }\) m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \left( \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } $$ $$ \therefore F = \left( \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \right) \left( \frac { \left( -1.6 \times { 10 }^{ -19 } \right) \times \left( +1.6 \times { 10 }^{ -19 } \right) }{ { \left( { 10 }^{ -9 } \right) }^{ 2 } } \right) $$ $$ \therefore F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \left( -2.56 \times { 10 }^{ -20 } \right) $$ $$ \therefore F = -23.063 \times { 10 }^{ -11 } N $$ The negative sign indicates that the force is attractive in nature.

Coulomb performed a number of experiments to see the effect of placing two small charges near each other. On his experimental researches, he established a law which is known as Coulombs Law of Electrostatics.

According to Coulombs Law :The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges \({ q }_{ 1 } \) and \({ q }_{ 2 }\) have distance d between them, then force F between the charges can be mathematically expressed as: $$ F \propto { q }_{ 1 }{ q }_{ 2 }\qquad ...(1)$$ $$ F \propto \frac { 1 }{ { d }^{ 2 } } \qquad ...(2)$$ Combining (1) and (2), $$ F \propto \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } or F = K\frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \qquad ...(3) $$ Where, \(K\) is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used.

In SI units force is measured in newton, charge in coulomb, distance in metre and the value of \(K\) is given as, $$ K = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \qquad ...(4) (in SI system)$$ where, \({ \epsilon }_{ 0 }\) = Absolute permittivity of vacuum or permittivity of free space = \(8.854 \times { 10 }^{ -12 }\) farad/metre
\( { \epsilon }_{ r }\) = Relative permittivity of the medium w.r.t. vacuum in which the charges are placed (e.g. for air \({ \epsilon }_{ r }\) = 1)

Putting the value of \(K\) in equation (3), we get, $$ \boxed { F = \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } \qquad ... (5)$$ Unit Charge or one coulomb charge in SI system can be defined as the amount of charge which when placed at a distance of one metre from an equal and similar charge in air, is repelled with a force of \(9 \times { 10 }^{ 9 } \)newton from it.

As per this defination, $$ { q }_{ 1 } = { q }_{ 2 } = q, $$ $$F = 9 \times { 10 }^{ 9 } newton,$$ $$ { \epsilon }_{ 0 } = 8.854 \times { 10 }^{ -12 } farad/metre,$$ \({ \epsilon }_{ r } = 1\) (Assuming two charges to be placed in air)
d = 1m

Putting these values in the above relation, we get $$ 9 \times { 10 }^{ 9 } = \frac { q \times q }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 \times { 1 }^{ 2 } } $$ $$ 9 \times { 10 }^{ 9 } = \frac { { q }^{ 2 } }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) } $$ $$ { q }^{ 2 } = \left( 9 \times { 10 }^{ 9 } \right) \times 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) = 1.00086 $$ $$ \therefore q = \pm 1.00043 \ coulomb $$ Hence, unit charge.

Determine the force between two charges, each of one coulomb when they are separated at one metre distance in air.

Solution:We have, $${ q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = 1m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \frac { 1 \times 1 }{ { \left( 1 \right) }^{ 2 } } $$ $$ F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } $$ $$ F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } $$ $$ F = 8.992 \times { 10 }^{ 9 } N $$

An electron and a proton are at a distance of 10^-9 m from each other in a free space. Compute the force between them.

Solution:We have, The charge on an electron, \({ q }_{ 1 } = -1.6 \times { 10 }^{ -19 } \) C
and the charge on a proton, \({ q }_{ 2 } = +1.6 \times { 10 }^{ -19 }\) C. $$ { q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = \({ 10 }^{ -9 }\) m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \left( \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } $$ $$ \therefore F = \left( \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \right) \left( \frac { \left( -1.6 \times { 10 }^{ -19 } \right) \times \left( +1.6 \times { 10 }^{ -19 } \right) }{ { \left( { 10 }^{ -9 } \right) }^{ 2 } } \right) $$ $$ \therefore F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \left( -2.56 \times { 10 }^{ -20 } \right) $$ $$ \therefore F = -23.063 \times { 10 }^{ -11 } N $$ The negative sign indicates that the force is attractive in nature.

The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as a whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. However, if from such a neutral body, electrons are removed, there occurs a shortage of electrons in the body.

Consequently the body no longer remains neutral. The result is that the body attains positive charge.

Thus, when a body is having shortage of electrons, it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge on the body.

To give a negative charge to any body, extra electrons must be supplied to it.

To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy.

This makes the charged body capable of doing work.

The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C).

1 coulomb of charge = The charge on \(625 \times {10}^{16}\) electrons.

Therefore, practical unit of charge is coulomb.

Like charges repel each other while unlike charges attract each other, means when two like charges are brought near to each other, they will repel each other and if two unlike charges [a +ve and a -ve charge] are brought near to each other, they will start attracting each other.

The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as a whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. However, if from such a neutral body, electrons are removed, there occurs a shortage of electrons in the body.

Consequently the body no longer remains neutral. The result is that the body attains positive charge.

Thus, when a body is having shortage of electrons, it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge on the body.

To give a negative charge to any body, extra electrons must be supplied to it.

To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy.

This makes the charged body capable of doing work.

The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C).

1 coulomb of charge = The charge on \(625 \times {10}^{16}\) electrons.

Therefore, practical unit of charge is coulomb.

Like charges repel each other while unlike charges attract each other, means when two like charges are brought near to each other, they will repel each other and if two unlike charges [a +ve and a -ve charge] are brought near to each other, they will start attracting each other.