Question

11. The mutual electrostatic force of attraction between two small spheres
carrying the charges +0.4 microcoulomb and - 0.8 microcoulomb in air is 0.2
Newton. What is the separation between two spheres.​

Answer 1

Given :

• Q 1 = 0.4 x 10^-6 C
• Q 2 = - 0.8 x 10^-6 C
• Electrostatic force (F) = 0.2 N

To find :

• Separation between the two spheres ,

How to solve :

• Substitute the given values in the formula in order to find the unkown variable

Solution :

Charges Q 1 and Q2 are attracted due to the electrostatic force acting between them is given by ,

$\leadsto F =\dfrac{ K Q_ 1 Q_ 2 }{r ^2}$

On rearranging ,

$\leadsto r ^2 =\dfrac{ K Q_ 1 Q_ 2 }{F}$

Now let's substitute the given values in the above equation ,

$\leadsto r ^2 =\dfrac{ 9 \times 10^9 \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6} }{0.2 }$

$\leadsto r ^2 = 9 \times 10^9 \times 0.4 \times 10^{-12} \times 4$

$\leadsto r ^2 = 14.4\times 10^{-3}$

$\leadsto r ^2 = 0.0144$

$\leadsto r = 0.12m$

Hence ,

The separation between the two spheres is 0.12 m .