Question

Q1=q , Q2 = -3q
F=-O.85N
r=28cm
q=?

According to Coloumb's law.​

Answer 1

given : Q1=q , Q2 = -3q

F=-O.85N

r=28cm

$= 28 \times {10}^{ - 2} m$

find

q=?

$f \: = \frac{1}{4\pi \: e \: o} \frac{q1 \: q2}{ {r}^{2} }$

$- 0.85 = \frac{9 \times {10}^{9} \times q \times ( - 3q) }{28 \times {10}^{ - 2} }$

${q}^{2} = \frac{ - 0.85(28 \times {10}^{ - 2) {}^{2} } }{9 \times {10}^{9} \times ( - 3q)}$

$= \frac{85 \times100 {}^{ - 1}28 \times 28 \times 10 {}^{ - 4} }{9 \times {10}^{9} \times - 3}$

$\frac{85 \times 28 \times 28 \times {10}^{12} }{27}$

therefore

$\frac{85 \times 28 \times 28 \times {10}^{12} }{27}$

$q = \sqrt{ \frac{85 \times 28 \times 28 \times {10}^{12} }{27} }$

aagye ga nhi horaha solve mere se I try .