Question

Given

Q1 = 0.4 x 10^-6 C
Q2 = -0.8 x 10^-6 C
F = 0.2 N
r = ?


Formula.
r^2 = 1/4π€°K . Q1 Q2 / F​

Answer 1

Answer:

the answer is 0.12m

Explanation:

$Q1 = 0.4 \times {10}^{ - 6}C \\ Q2 = - 0.8 \times {10}^{ - 6} C$

F = 0.2N

Applying the forumula of Coloumb's force of attraction or repulsion,

$F = \frac{ Q 1 \times Q 2}{ 4\pi ε0} \times \frac{1}{ {r}^{2} }$

Now we will take only the magnitude of charges i.e. we will not consider the '+','-' sign.

ε0 = permittivity of free space=

$ε0 = 8.85 \times {10}^{ - 12} { {N}^{ - 1} {m}^{ - 2} C}^{ 2}$

$\frac{ 1}{ 4\pi ε0} \ = 9 \times {10}^{9} {N {m}^{2} C}^{ - 2}$

$F = \frac{ Q 1 \times Q 2}{ 4\pi ε0} \times \frac{1}{ {r}^{2} } \\ {r }^{2} = \frac{ Q 1 \times Q 2}{ 4\pi ε0 \times \: F } \\ {r }^{2}= \frac{0.4 \times {10}^{ - 6 } \times 0.8 \times {10}^{ - 6} }{0.2} \times 9 \times {10}^{9} \\ {r }^{2}= \frac{0.32 \times {10}^{ - 12} }{0.2} \times 9 \times {10}^{9} \\ {r }^{2} = 16 \times {10}^{ - 13} \times 9 \times {10}^{9} \\ r= \sqrt{144 \times {10}^{ - 4}} \\ r= 12 \times {10}^{ - 2} = 0.12m$