Two copper balls, each weighing $10\,g$ are kept in air $10\, cm$ apart. If one electron from every ${10^6}$ atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is $63.5$)

  • A

    $2.0 \times {10^{10}}\,N$

  • B

    $2.0 \times {10^4}\,N$

  • C

    $2.0 \times {10^8}\,N$

  • D

    $2.0 \times {10^6}\,N$

Similar Questions

Two identical conducting spheres having unequal positive charges $q_1$ and $q_2$ separated by distance $r$. If they are made to touch each other and then separated again to the same distance, the electrostatic force between them in this case will be :-

Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart, they repel each other with a force $F$. A third conducting sphere identical to the other two, but initially uncharged is touched to one sphere and then to the other before being removed. The force between the original two spheres is now

  • [KVPY 2009]

Positive point charges are placed at the vertices of a star shape as shown in the figure. Direction of the electrostatic force on a negative point charge at the centre $O$ of the star is

  • [KVPY 2017]

Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. A third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now

  • [KVPY 2014]

There is another useful system of units, besides the $\mathrm{SI/MKS}$. A system, called the $\mathrm{CGS}$ (centimeter-gramsecond) system. In this system Coloumb’s law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm\left( { = {{10}^{ - 2}}m} \right)$ , $\mathrm{F}$ in dynes $\left( { = {{10}^{ - 5}}N} \right)$  and the charges in electrostatic units $(\mathrm{es\,unit}$), where $1$ $\mathrm{esu}$ of charge $ = \frac{1}{{[3]}} \times {10^{ - 9}}C$. The number ${[3]}$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times {10^8}m/s$. An approximate value of $c$ then is $c = 3 \times {10^8}m/s$.

$(i)$ Show that the coloumb law in $\mathrm{CGS}$ units yields $1$ $\mathrm{esu}$ of charge = $= 1\,(dyne)$ ${1/2}\,cm$. Obtain the dimensions of units of charge in terms of mass $\mathrm{M}$, length $\mathrm{L}$ and time $\mathrm{T}$. Show that it is given in terms of fractional powers of $\mathrm{M}$ and $\mathrm{L}$ .

$(ii)$ Write $1$ $\mathrm{esu}$ of charge $=xC$, where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi { \in _0}}} = \frac{{{{10}^{ - 9}}}}{{{x^2}}}\frac{{N{m^2}}}{{{C^2}}}$ with $x = \frac{1}{{[3]}} \times {10^{ - 9}}$ we have, $\frac{1}{{4\pi { \in _0}}} = {[3]^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ or $\frac{1}{{4\pi { \in _0}}} = {\left( {2.99792458} \right)^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ (exactly).