Question

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

Answer 1

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

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

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

According to Coulomb's law,

Force between two charged particles in air is given by:

$force = \frac {k(q1 \times q2)}{r}$

$k = \frac{1}{4\pi \: \: \: \: \: \: epsilon}$

= 9 x 10^9 N m² /C²

q1 and q2 = charges of two charged spheres

r = distance between charged spheres

Force between between two small charged spheres charges kept apart in air =

(9 x 10^9 N m² /C²) * (2 x 10^(-7) C) * (3 x 10^(-7) C)/(30 x 10^-2 m)²

= 3/500

= 6 x 10^-3 N

or 6 mN.

Nature of force = Repulsive.

So Force => 6 mN Repulsive.

@HarshPratapSingh

Answer 2

$\sf\huge{\green{\underline{\underline{\orange{ANSWER}}}}}$

$\sf{\huge\bold{\underline{\underline{\red{Given:-}}}}}$

↓↓↓↓ ${Two\:small\:charged\:sphere\:having\:charges\:of\:,}$

• ${q_1}$ = 2×10$^{-7}$ C
• ${q_2}$ = 3×10$^{-7}$ C
• “r”(distance between them) = 30 cm = 0.3 m

${\huge{\bold{\purple{\bigstar{\boxed{Formula:-}}}}}}$

• Force = ${\dfrac{1}{4\:{\pi}\:{\epsilon_o}}}$ × ${\dfrac{q_1\:q_2}{r^2}}$

☞we have know that,

• ${\dfrac{1}{4\:{\pi}\:{\epsilon_o}}}$ = 9×10$^{9}$ N.m²/C²

$\bigstar$ $\huge\purple{\bf{\underline{SOLUTION}}}$

☞ Now put the formula and get the Answer;

$\tt{\implies\:{F\:=\:(9\:\times\:10{^9})\:\times\:{\dfrac{(2\:\times\:10^{-7})\:\times\:(3\:\times\:10^{-7})}{(0.3)^2}}}}$

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

✔️As $q_1$ and $q_2$ are positive charges , the force between them is repulsive .

$\bigstar\:\underline{\boxed{\bf{\red{Required\: Answer\:=\:6\:\times\:10^{-3}\:N}}}}$