Question

Two point changes, q, = 5x 106 and
q2 =3x106 & the force
between them 180N
Calculate the distance between
them​

Answer 1

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⭐ Two point changes, 5 × ${10}^{6}$

and 3 × ${10}^{6}$ & the force between them 180N . Calculate the distance between them.

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

➡ 8.67m²

$\mathtt{\huge{\underline{\orange{Solution:-}}}}$

Given :-

• q1 = 5 × ${10}^{6}$

• q2 =3 × ${10}^{6}$

• Force = 180N

To Find :-

• The distance between them.

Concept :-

Columbs Law state that the force of attraction or repulsion between the two stationary point charge is

• Directly proportional to the product of magnitude of two charges.

• Inversely proportional to the square of distance between them.

Calculation :-

According to the columbs law ,

Force = $= \dfrac{1}{4πɛ} \times \dfrac{q1q2}{r²}$

➝ 180 N = 1/4πɛ × 3×5 ×10^12 / r²

➝ 180N = 9 × ${10}^{-9}$ × $\dfrac{15 \times {10}^{12} }{r ^{2} }$

➝ 180 N =9 × 15 × ${10}^{3}$ / r²

➝ 180 N =135 × ${10}^{3}$ / r²

➝ $\dfrac{180}{135 \times {10}^{3} } = \dfrac{1}{r ^{2} }$

➝ ${r}^{2} = \dfrac{135 \times {10}^{3} }{180}$

➝ r² = 0.75 × ${10}^{3}$

➝ r² = 75 ×10

➝ r = $\sqrt{750 }$

➝ r = 8.67m²

So, the distance sphere is 8.67m².

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