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>>Laws of Motion
>>Types of Forces
>> $Non-Contact Forces$

Non-Contact Forces

Q: F_g and F_e represent gravitational and electrostatic force respectively between electrons situated at a distance 0.1\,m. F_g/F_e is of the order

{ F }_{ y } and { F }_{ e } represent gravitational and electrostatic force respectively between electrons situated at a distance =0.1m let, us take the ratio \dfrac { { F }_{ e } }{ { F }_{ g } } =\dfrac { \dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }^{ 2 } }{ { d }^{ 2 } } }{ G\dfrac { { m }^{ 2 } }{ { d }^{ 2 } } } \dfrac { { F }_{ e } }{ { F }_{ g } } =\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \times \dfrac { 1 }{ G } \times \dfrac { { q }^{ 2 } }{ { m }^{ 2 } } \dfrac { { F }_{ e } }{ { F }_{ g } } =9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 6.67\times { 10 }^{ -11 } } \times \dfrac { { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ { \left( 9.1\times { 10 }^{ -31 } \right) }^{ 2 } } \Rightarrow \dfrac { { F }_{ e } }{ { F }_{ g } } =4.17\times { 10 }^{ 42 } or, \dfrac { { F }_{ e } }{ { F }_{ g } } is of the order of { 10 }^{ -42 } .

Q: The ratio of magnitude of electrostatic force and gravitational force for an electron and proton is

Here, for an electron and a protonElectrostatic force, \left| { F }_{ e } \right| =\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { e }^{ 2 } }{ { r }^{ 2 } } Gravitational force, \left| { F }_{ g } \right| =\cfrac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \therefore \cfrac { \left| { F }_{ e } \right| }{ \left| { F }_{ g } \right| } =\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { e }^{ 2 } }{ G{ m }_{ e }{ m } } =\cfrac { 9\times { 10 }^{ 9 }\times { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ 6.67\times { 10 }^{ -11 }\times 9\times { 10 }^{ -31 }\times 1.66\times { 10 }^{ -27 } } =2.4\times { 10 }^{ 39 }

Q: Differentiate between contact and non-contact force. Support your answer with suitable examples.

Contact forces Non-contact forces These are forces which need a physical touch on the object to act upon it. These are forces which do not need a physical contact to act on an object. These may or may not be natural forces. These are natural forces They can not work from a distance. They can work from a distance as well. Its various types are applied force, normal force, tension frictional force its various types are magnetic force electrostatic force,gravitational force Examples player kiicking a ball game of tug of war rolling ball stops itself horse pulling a cart etc, Examples: magnet attracting iron nails, mango falling from tree, charged plastic ruler attracts bits of dry - paper etc.

Q: When two equal forces act on an object in opposite directions, they are known as:

Balanced forces are equal and opposite in direction, they act on the same body. The net force due to these forces is zero.

Q: Check that the ratio ke^2 /G m_e\,m_p is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

The given ratio is \displaystyle \frac{ke^2}{Gm_cm_p} Where, G = Gravitational constantIts unit is N\,m^2kg^{-2} . m_e and m_p= Masses of electron and proton.Their unit is kg. e = Electric charge.Its unit is C. k = A constant \displaystyle =\frac{1}{4\pi\in_0} \in_0= Permittivity of free spaceIts unit is N\,m^2C^{-2} . Unit of the given ratio \displaystyle \frac{ke^2}{Gm_cm_p}=\frac{[Nm^2C^2][C^{-2}]}{[Nm^2\,kg^{-2}][kg][kg]} =M^0L^0T^0 Hence, the given ratio is dimensionless. e=1.6\times 10^{-19}C G=6.67\times 10^{-11}N\,m^2kg^{-2} m_e=9.1\times 106{-31}kg m_p=1.66\times 10^{-27}kg \displaystyle \frac{ke^2}{Gm_cm_p} = \frac{9\times 10^9\times(1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-3}\times 1.67\times 10^{-22}}\approx 2.3\times 10^{39} This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Q: The ratio of electric force to the gravitational force between two protons separated by a finite distance is of the order of:

Gravitational\ force\ = \dfrac{G(m_p)^2}{r^2} Electric\ force\ = \dfrac{K(e)(e)}{r^2} Ratio\ = \dfrac{K(1.6\times 10^{-19})^2}{r^2}\times \dfrac{r^2}{G\times (1.6\ 7\times 10^{-27})^2} = \dfrac{(9\times 10^9)\times (1.6\times 10^{-19})^2}{(6.6\ 7\times 10^{-11})(1.6\ 7\times 10^{-27})^2} =1.23\times 10^{20}\times 10^{16} = 10^{36} (order of magnitude)

Q: Arrange the following basic forces in the increasing order of relative strength.1. Gravitational force2. Electromagnetic force3. Weak nuclear force4. Strong nuclear force

Strength wise order :Strong nuclear force > Electromagnetic force > Weak nuclear force > gravitational force. 4>2>3>1 .

Q: When a body is stationary :

For the body to be stationary, net external force acting on it must be equal to zero which is possible only when the combination of forces acting on it balance each other.

Q: Like poles of two magnets _____ each other.

Like poles of two magnets repel each other, The force between them is non contact force and magnetic force

Q: Compute the ratio of the gravitational force and electrostatic force between an electron (mass 9.1\times {10}^{31}kg ) and a proton (mass 1.7\times {10}^{-27}kg )

Mass of electron =9.1\times 10^{-31} (value given in the question is wrong)Let us assume that electron and proton placed with separation distance r meters.Gravitational Force F_G=\dfrac{G\times m_p\times m_e}{r^2}=\dfrac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.7\times 10^{-27}}{r^2}=\dfrac{6.67\times 9.1\times 1.7\times 10^{-69}}{r^2}.....................(i) Electrostatic Force F_E=\dfrac{1}{4\pi \varepsilon_0} \dfrac{e^2}{r^2}=\dfrac{9\times 10^9\times 1.6\times 1.6\times 10^{-38}}{r^2}.................................................(ii) Hence, \dfrac {F_G}{F_E}=4.47\times 10^{-40}

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