Question

Two point charges 8 by 3 Cand 2 by 3 C are kept in a vacuum. What is the distance between them if they repel each other with a force of 1.6 ×10*12 N ?

Answer 1

Answer:

$\displaystyle \text{r = 0.1 m}$

Explanation:

$\displaystyle \text{Given}\\\\\\\displaystyle \text{ q_1= \dfrac{8}{3} C and q_2=\dfrac{2}{3} C }\\\\\\\displaystyle \text{Force = 1.6\times10^{12} N}\\\\\\\displaystyle \text{We know formula for Force}\\\\\\\displaystyle F=\dfrac{1}{4\pi\epsilon}\dfrac{q_1q_2}{r^2} \\\\\\\displaystyle \text{We know value of : }\\\\\\\displaystyle F=\dfrac{1}{4\pi\epsilon}=9.0\times10^{9}\\\\\\\displaystyle \text{Now putting given values}$

$\displaystyle{1.6\times10^{12}=9.0\times10^{9}\times\dfrac{8/3\times2/3}{r^2} }\\\\\\\displaystyle{1.6\times10^{3}=9.0\times\dfrac{16/9}{r^2}}\\\\\\\displaystyle{r^2=9.0\times\dfrac{16/9}{1.6\times10^3}}\\\\\\\displaystyle{r^2=1\times10^{-2}}\\\\\\\displaystyle{r=10^{-1}}\\\\\\\displaystyle \text{r = 0.1 m}$

$\displaystyle \text{Hence we get answer .}$

Answer 2

Answer:

${\bold{\therefore r=0.1\:m}}$

$\huge{\bold{\underline{\underline{SOLUTION-}}}}$

• In the given question information given about charges of two point and force is given.

• We have to find distance between the two points.

$\underline \bold{Given : } \\ \implies q_{1} = \frac{8}{3} c \\ \\ \implies q_{2} = \frac{2}{3} c \\ \\ \implies F = 1.6 \times {10}^{12} n \\ \\ \underline \bold{To \: Find : } \\ \implies r = ?$

• According to given question :

$\bold {By \: using \: formula : } \\ \implies F= \frac{1}{4\pi \in} \frac{q _{1} q_{2} }{ {r}^{2} } \\ \\ \bold{where, } \\ \implies \frac{1}{1\pi \in} = 9.0 \times {10}^{9} \\ \\ \bold{putting \: given \: values} \\ \implies 1.6 \times {10}^{12} = 9.0 \times {10}^{9} \times \frac{ \frac{8}{3} \times \frac{2}{3} }{ {r}^{2} } \\ \\ \implies 1.6 \times {10}^{12} = 9.0 \times {10}^{9} \times \frac{ \frac{16}{9} }{ {r}^{2} } \\ \\ \implies \frac{1.6 \times {10}^{12} }{9.0 \times {10}^{9} } = \frac{16}{9( {r})^{2} } \\ \\ \implies \frac{16 \times {10}^{11 - 9} }{9.0} = \frac{16}{9( {r})^{2} } \\ \\ \implies \frac{ \cancel{16 }\times {10}^{2} \times \cancel9 }{ \cancel{9} \times \cancel{16}} = \frac{1}{ {r}^{2} } \\ \\ \implies {r}^{\cancel2} = \frac{1}{ {10}^{\cancel2} } \\ \\ \implies r = \frac{1}{10} \\ \\ \bold{\implies r = 0.1 \: m}$