Electric Charges and Fields

Physics Part I

Physics

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The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μc in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

(a) Given,

$\mathrm{Force}-\mathrm{F}=0.2\mathrm{N}$

$\mathrm{charge}-{\mathrm{q}}_{1}=0.4\mathrm{\mu C}=0.4\times {10}^{-6}\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{charge}-{\mathrm{q}}_{2}=0.8\mathrm{\mu C}=0.8\times {10}^{-6}\mathrm{C}$

Now using the formula, $\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}$

Hence, $\overline{){\mathrm{r}}^{2}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{\mathrm{F}}}$

${\mathrm{r}}^{2}=\frac{9\times {10}^{9}\times 0.4\times {10}^{-6}\times 0.8\times {10}^{-6}}{0.2}\phantom{\rule{0ex}{0ex}}{\mathrm{r}}^{2}=36\times 4\times {10}^{-4}=144\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\mathrm{r}=12\times {10}^{-2}\mathrm{m}=12\mathrm{cm}.$

where, r is the distance between two spheres.

(b) Force on the second sphere due to the first is same, i.e., 0.2 N because the charges in action are same and force is attractive as charges are unlike in nature.

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a) Explain the meaning of the statement 'electric charge of a body is quantised'.

b) why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

b) why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

a) Quantisation of Electric Charges mean the total electric charge(q) of a body is always an integral multiple of a basic quantum charge(e).

i.e., $\mathrm{q}=\pm \mathrm{ne}$

Here +e is taken as charge on a proton while –e is taken as charge on an electron. The charge on a proton and an electron are numerically equal i.e., 1.6 x 10

“Quantisation is a property due to which charge exists in discrete packets in multiple of

± 1.6 x 10^{–19} rather than in continuous amounts.”

b) Based on many practical phenomena, we may ignore quantisation of electric charge and consider the charge to be continuous. In a macroscopic scale the number of charges used is enormous as compared to the magnitude of charge. The “graininess” of charge is lost and it appears continuous and therefore quantisation of charge becomes insignificant.

b) Based on many practical phenomena, we may ignore quantisation of electric charge and consider the charge to be continuous. In a macroscopic scale the number of charges used is enormous as compared to the magnitude of charge. The “graininess” of charge is lost and it appears continuous and therefore quantisation of charge becomes insignificant.

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When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Law of conservation of charge states that total charge on an isolated system of objects always remain conserved.

When a glass rod is rubbed with silk cloth, glass rod becomes positively charged while silk cloth becomes negatively charged and, the amount of positive charge on the glass rod is apparently found to be exactly the same as the negative charge on silk cloth. Since, the measure of charge is same on both, equal amount of charge with opposite nature will cancel out each other. Hence, the total sum of charge on two bodies is zero. Thus, the system of glass rod and silk cloth, which was neutral before rubbing, still possesses no net charge after rubbing. And law of conservation of charge is justified.

As a consequence of conservation of charge, when two charged conductors of same size and same material carrying charges Q_{1} and Q_{2} respectively are brought in contact and separated, the charge on each conductor will be $\frac{{\mathrm{Q}}_{1}+{\mathrm{Q}}_{2}}{2}.$

This condition, however, does not hold true if the conductors are of different sizes or of different material.

In that case the charges on the conductors will be Q_{1}^{' }and Q_{2}’ respectively, where Q_{1} + Q_{2} = Q_{1}‘ + Q_{2}’.

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Check that the ratio $\frac{{\mathrm{ke}}^{2}}{{\mathrm{Gm}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}}$is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

$\mathrm{Electrostatic}\mathrm{force}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{F}=\mathrm{K}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Gravitational}\mathrm{force}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{F}=\mathrm{G}\frac{{\mathrm{m}}_{1}{\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}$

$\mathrm{where}\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}{\mathrm{m}}^{2}{\mathrm{kg}}^{-2}\phantom{\rule{0ex}{0ex}}{\mathrm{m}}_{\mathrm{e}}=9.1\times {10}^{-31}\mathrm{kg}\phantom{\rule{0ex}{0ex}}{\mathrm{m}}_{\mathrm{p}}=1.67\times {10}^{-27}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$

and $\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}=9\times {10}^{9}\frac{{\mathrm{Nm}}^{2}}{{\mathrm{C}}^{2}}$

Now, $\frac{{\mathrm{ke}}^{2}}{{\mathrm{Gm}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}}=\frac{9\times {10}^{9}{\displaystyle \frac{{\mathrm{Nm}}^{2}}{{\mathrm{C}}^{2}}}\times 1.6\times {10}^{-19}\mathrm{C}\times 1.6\times {10}^{-19}\mathrm{C}}{6.67\times {10}^{-11}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{-2}\times 9.1\times {10}^{-31}\mathrm{kg}\times 1.67\times {10}^{-27}\mathrm{kg}}\phantom{\rule{0ex}{0ex}}=2\times 27\times {10}^{39}\mathrm{which}\mathrm{is}\mathrm{dimensionless}.$

It also establishes that the electrostatic force is about 10^{39} times stronger than the gravitational force.

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Given,

$C\mathrm{harge},{q}_{1}=2\times {10}^{-7}\mathrm{C},\phantom{\rule{0ex}{0ex}}C\mathrm{harge},{\mathrm{q}}_{2}=3\times {10}^{-7}\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{r}=30\mathrm{cm}=0.3\mathrm{m}$

where, r is the distance between two charges.

Using the formula,

$\therefore $ $\overline{)\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}}$

$=\frac{9\times {10}^{9}\times 2\times {10}^{-7}\times 3\times {10}^{-7}}{(0.3{)}^{2}}\phantom{\rule{0ex}{0ex}}=6\times {10}^{-3}\mathrm{N}\left(\mathrm{Repulsive}\right)$

Repulsive in nature since both charges are positive.

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