The force of attraction between two charges 8 μ C and -4 μ C is 0.2 N . Find the distance of separation

Correct answer: A. 1.2 m Given that, F=0.2 ~N, Q_1=8 μ C, Q_2=4 μ C Force between two charges, F=k Q_1 Q_2 r^2 or 0.2=9 × 10^9 × 8 × 10^ -6 × 4 × 10^ -6 r^2 r^2= 9 × 10^9 × 8 × 10^ -6 × 4 × 10^ -6 0.2 or r=1.2 ~m

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Electrostatics

JIPMER 2018 · Physics · Electrostatics

The force of attraction between two charges 8 μ C and -4 μ C is 0.2...

The force of attraction between two charges

8 \mu \mathrm{C}

and

-4 \mu \mathrm{C}

is 0.2 N . Find the distance of separation

Options

A $1.2 m$
B $12 m$
C $120 m$
D $0.12 m$

Correct answer

A. $1.2 m$

Step-by-step solution

Given that,

F=0.2 \mathrm{~N}, \mathrm{Q}_1=8 \mu \mathrm{C}, \mathrm{Q}_2=4 \mu \mathrm{C}

\because

Force between two charges,

\mathrm{F}=\mathrm{k} \frac{\mathrm{Q}_1 \mathrm{Q}_2}{\mathrm{r}^2}

\quad 0.2=9 \times 10^9 \times \frac{8 \times 10^{-6} \times 4 \times 10^{-6}}{\mathrm{r}^2}

\therefore \quad r^2=\frac{9 \times 10^9 \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{0.2}

\quad \mathrm{r}=1.2 \mathrm{~m}

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