Question

What is the force between two small charged spheres having charges of 2 x 10^-7c and 3 x 10^-7c placed 30 cm in air.

Answer 1

Heya mate....

Charge on the first sphere, q1 = 2 × 10^ - 7 C

Charge on the second sphere, q2 = 3 × 10 ^- 7 C

Distance between the spheres, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

F=q1 q2/4 pi E• r^2

Where, E•= Permittivity of free space

F=9×19^9 × 2×3×10^-14÷0.3×0.3

=6×10^-3N

Hence, force between the two small charged spheres is 6 × 10 ^- 3 N. The charges are of same nature. Hence, force between them will be repulsive.

@skb

Answer 2

$\textbf{ Hey there}$!


$\textbf{Solution}$:


Here $q_{1}$ = 2 × $10^{-7}$


And $q_{2}$ = 3 × $10^{-7}$


Distance ( r ) = 30 cm.


When it is converted in m we get:

= 0.3 m


The Force in air, using coulomb's law:


F = $\frac{1}{4E_{0}\pi}$.$\frac{q1q2}{r^{2}}$


We know that the value of 1/4πE0 = 9 × $10^{9}$


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


= 6 × $10^{-3}$ N.


Where N is Newton.


#Be Brainly.