Two charges each of $1\;coulomb$ are at a distance $1\,km$ apart, the force between them is
Two charges each of $1\;coulomb$ are at a distance $1\,km$ apart, the force between them is
- A
$9 \times {10^3}\;Newton$
- B
$9 \times {10^{ - 3}}\;Newton$
- C
$1.1 \times {10^{ - 4}}\;Newton$
- D
${10^4}\;Newton$
Similar Questions
Two particle of equal mass $m$ and charge $q$ are placed at a distance of $16\, cm$. They do not experience any force. The value of $\frac{q}{m}$ is
Two particle of equal mass $m$ and charge $q$ are placed at a distance of $16\, cm$. They do not experience any force. The value of $\frac{q}{m}$ is
Two particles $X $ and $Y$, of equal mass and with unequal positive charges, are free to move and are initially far away from each other. With $Y$ at rest, $X$ begins to move towards it with initial velocity $u$. After a long time, finally
Two particles $X $ and $Y$, of equal mass and with unequal positive charges, are free to move and are initially far away from each other. With $Y$ at rest, $X$ begins to move towards it with initial velocity $u$. After a long time, finally
Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.
$(a)$ Compare the strength of these forces by determining the ratio of their magnitudes $(i)$ for an electron and a proton and $(ii)$ for two protons.
$(b)$ Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \mathring A \left( { = {{10}^{ - 10}}m} \right)$ apart? $\left(m_{p}=1.67 \times 10^{-27} \,kg , m_{e}=9.11 \times 10^{-31}\, kg \right)$
Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively.
$(a)$ Compare the strength of these forces by determining the ratio of their magnitudes $(i)$ for an electron and a proton and $(ii)$ for two protons.
$(b)$ Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \mathring A \left( { = {{10}^{ - 10}}m} \right)$ apart? $\left(m_{p}=1.67 \times 10^{-27} \,kg , m_{e}=9.11 \times 10^{-31}\, kg \right)$
Two identical conducting spheres $\mathrm{P}$ and $\mathrm{S}$ with charge $Q$ on each, repel each other with a force $16 \mathrm{~N}$. A third identical uncharged conducting sphere $\mathrm{R}$ is successively brought in contact with the two spheres. The new force of repulsion between $\mathrm{P}$ and $\mathrm{S}$ is :
Two identical conducting spheres $\mathrm{P}$ and $\mathrm{S}$ with charge $Q$ on each, repel each other with a force $16 \mathrm{~N}$. A third identical uncharged conducting sphere $\mathrm{R}$ is successively brought in contact with the two spheres. The new force of repulsion between $\mathrm{P}$ and $\mathrm{S}$ is :
- [JEE MAIN 2024]
Four point charges, each of $+ q$, are rigidly fixed at the four corners of a square planar soap film of side ' $a$ ' The surface tension of the soap film is $\gamma$. The system of charges and planar film are in equilibrium, and $a=k\left[\frac{q^2}{\gamma}\right]^{1 / N}$, where ' $k$ ' is a constant. Then $N$ is
Four point charges, each of $+ q$, are rigidly fixed at the four corners of a square planar soap film of side ' $a$ ' The surface tension of the soap film is $\gamma$. The system of charges and planar film are in equilibrium, and $a=k\left[\frac{q^2}{\gamma}\right]^{1 / N}$, where ' $k$ ' is a constant. Then $N$ is
- [IIT 2011]