Question

On a sphere of 0.5µc charge, 0.2 N force is applied to the air due to any other small charged sphere! If the other sphere charge -0.8µc, then how much distance between the two spheres.​

Answer 1

Answer:

• The Distance (r) between the spheres is 0.1314 m

Given:

1. Charge (Q₁) = 0.5 µ C
2. Charge (Q₂) = - 0.8 µ C
3. Electrostatic Force (F) = 0.2 N

Explanation:

$\rule{300}{1.5}$

From the formula,

$\large{\bigstar\;\boxed{\tt F = \dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{Q_1\;Q_2}{r^2}}}$

$\frak{here}\begin{cases}\sf{Q_1}\text{ Denotes First charge}\\\sf{Q_2}\text{ Denotes Second charge}\\\text{r Denotes Distance}\end{cases}$

Now,

$\large{\boxed{\tt F = \dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{Q_1\;Q_2}{r^2}}}$

Note,

we have to take the magnitude of the charges i.e. two charges will be |Q₁| = 0.5μC and, |Q₂| = 0.8μC

Converting the charges into S.I units,

Q₁ = 0.5 μC = 0.5 × 10⁻⁶ C Q₂ = 0.8 μC = 0.8 × 10⁻⁶ C

Substituting the values,

$\displaystyle \dashrightarrow\tt 0.2=\dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{0.5\times10^{-6}\times 0.8\times10^{-6} }{r^2}\\\\\\\dashrightarrow\tt 0.2=\dfrac{1}{4 \pi \epsilon_0}\;.\;\dfrac{0.4\times 10^{-6-6}}{r^2}\\\\\\\dashrightarrow\tt 0.2=9\times 10^9\;.\;\dfrac{0.4\times 10^{-12}}{r^2}\\\\\\\dag\tt \qquad \dfrac{1}{4\pi \epsilon_0}=9\times 10^9\\\\\\\dashrightarrow\tt 0.2\times r^2=9\times0.4\times 10^{9-12}$

$\displaystyle \dashrightarrow\tt 0.2\times r^2=3.6\times 10^{-3}\\\\\\\dashrightarrow\tt r^2=\dfrac{3.6\times 10^{-3}}{0.2}\\\\\\\dashrightarrow\tt r^2=18\times10^{-3}\\\\\\\dashrightarrow\tt r^2=1.8\times 10^{-2}\\\\\\\dashrightarrow\tt r=\sqrt{1.8\times 10^{-2}}\\\\\\\dashrightarrow\tt r=1.314\times0.1\\\\\\\dashrightarrow \large{\underline{\boxed{\red{\tt r=0.1314\;m}}}}$

∴ The Distance (r) between the spheres is 0.1314 m.

$\rule{300}{1.5}$