Question

What is the force between two small charged spheres having charges of 2x10⁻⁷ C and 3x10⁻⁷ placed 30Cm apart in air?

Answer 1

Answer:

$F\ =\ 6\times 10^{-3}\ N$

Explanation:

Given,

* $q_1\ =\ 2\times 10^{-7}\ C$

*$q_2\ =\ 3\times 10^{-7}\ C$

* Distance between the charges = r = 30 cm = 0.30 m

Therefore,

Electric force

$F\ =\ \dfrac{q_1q_2}{4\pi \epsilon_0 r^2}\\\Rightarrow F\ =\ \dfrac{9\times 10^9\times 2\times 10^{-7}\times 3\times 10^{-7}}{0.3^2}\\\Rightarrow F\ =\ 6\times 10^{-3}\ N$

Answer 2

Answer:

Force = $6\times 10^{-3}} N$

Explanation:

Given:

         charge on first sphere, $q_{1} = 2\times 10^{-7} C$

         charge on other sphere, $q_{2} = 3\times 10^{-7} C$

 distance between two charges = 30 cm = 0.3 m ∵ (1m = 100 cm)

Now ,

force between two charged particles is given by ,

   $F= \frac{k q_{1}q_{2}}{r^{2}}$     ,where $k=9\times 10^{9} N$

   $F=\frac{9\times 10^{9}\times 2\times 10^{-7}\times 3\times 10^{-7}}{\left ( 0.3 \right )^{2}}$

   $F=6\times 10^{-3}} N$