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>>Class 11
>>Physics
>>Laws of Motion
>>Types of Forces
>> $Non-Contact Forces$

Non-Contact Forces

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: \$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 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: 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: 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: 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 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: Two particles each of mass \$m\$ and carrying charge \$Q\$ are separated by some distance. If they are in equilibrium under mutual gravitational and electrostatic forces, then Q/m (in C/Kg) is of the order of:

Given : Gravitational force = Electrostatics forceLet the distance between the rwo charges be \$r\$.\$\Rightarrow \dfrac{G(m)m}{r^2}= \dfrac{K\ Q\ . Q}{r^2}\$\$\Rightarrow \dfrac{G}{K}= \dfrac{Q^2}{m^2}\$\$\Rightarrow\sqrt{\dfrac{6.67\times 10^{-11}}{9\times 10^9}}= \dfrac{Q}{m}\$\$\therefore \dfrac{Q}{m}\$ is the order of \$10^{-10}\$

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: Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

\$\textbf {Hint :}\$ When equal and opposite forces act on any object, then they cancel out each other so that the total external force experienced by the object is zero. \$\textbf{Step1: Draw free body diagram}\$\$\textbf{Step2: Calculate net force}\$So net force, \$F=m\times a\$Where, F=Net force, m=Mass, a=Acceleration.Hence, If net force (F) =0, then acceleration (a) will also be equals to zero (because mass ‘m’ cannot be zero). Therefore, acceleration in an object could be zero even when several forces are acting on it. Hence, Agree with Soni.

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