Question

Helppppo!!!

Answer 1

Answer:

Electric force = $3.15\times 10^{-11}\textrm{ N}$.

Explanation:

Given:

Charge on one grain, $q_{1}=6\times 10^{-10}\textrm{ C}$

Charge on second grain, $q_{2}=2.3\times 10^{-15}\textrm{ C}$

Distance between the two grains is, $r=2\textrm{cm}=0.02\textrm{ m}$

Electric force acting between two charges $q_{1}\textrm{ and }q_{2}$ separated by a distance $r$ is:

$F_{e}=\frac{kq_{1}q_{2}}{r^2}$

Where, $k$ is Coulomb's constant equal to $9\times 10^9\textrm{ }Nm^2/C^2$.

Now, plug in the given values and solve for $F{e}$.

$F{e}=\frac{9\times 10^9\times 6.0\times 10^{-10}\times 2.3\times 10^{-15}}{(0.02)^2}\\\\F_{elec}=3.15\times 10^{-11}\textrm{ N}$

Therefore, the electric force between them is $3.15\times 10^{-11}\textrm{ N}$.