Question

A charge of 5 x10-9 c is attracted by a charge of -3x10-7 with a force of 0.135 n

Answer 1

Question :

A charge of 5 × 10-⁹ C is attracted by a charge of - 3 × 10-⁷C with a force that 0.135 N , find the distance between them.

Given :

• Charge (Q) = 5 × 10-⁹C.

• Charge (q) = -3 × 10-⁷C.

• Force (F) = 0.135 N.

To find :

Distance between the two charges.

Solution :

We know the formula for Force between two charges i.e,

$\boxed{\bf{F = \dfrac{1}{4\pi \epsilon_{0}}\:\dfrac{Qq}{r^{2}}}}$

Where :

• F = Force field .
• Q = Charge on the body.
• q = Charge on the body.
• r = Distance between the two charges.
• 1/4πε = E = Electric field strength.

Now using the above formula and substituting the values in it, we get :

$:\implies \bf{F = \dfrac{1}{4\pi \epsilon_{0}}\:\dfrac{Qq}{r^{2}}}$

$:\implies \bf{0.135 = 9 \times 10^{9} \times \dfrac{5 \times 10^{-9} \times (-3) \times 10^{-5}}{r^{2}}}$

$:\implies \bf{0.135 = 9 \times 10^{9} \times \dfrac{5 \times 10^{-9} \times 3 \times 10^{-5}}{r^{2}}}$

$:\implies \bf{0.135 = 27 \times 10^{9} \times \dfrac{5 \times 10^{(-9) + (-5)}}{r^{2}}}$

$:\implies \bf{0.135 = 27 \times 10^{9} \times \dfrac{5 \times 10^{-14}}{r^{2}}}$

$:\implies \bf{0.135 \times r^{2} = 27 \times 10^{9} \times 5 \times 10^{-14}}$

$:\implies \bf{0.135 \times r^{2} = 135 \times 10^{-14 + 9}}$

$:\implies \bf{0.135 \times r^{2} = 135 \times 10^{-5}}$

$:\implies \bf{r^{2} = \dfrac{135 \times 10^{-5}}{0.135}}$

$:\implies \bf{r^{2} = \dfrac{135 \times 10^{-5}}{\dfrac{135}{1000}}}$

$:\implies \bf{r^{2} = \dfrac{135 \times 10^{-5} \times 10^{3}}{135}}$

$:\implies \bf{r^{2} = 10^{-5} \times 10^{3}}$

$:\implies \bf{r^{2} = 10^{-2}}$

$:\implies \bf{r^{2} = \dfrac{1}{10^{2}}}$

$:\implies \bf{r = \sqrt{\dfrac{1}{10^{2}}}}$

$:\implies \bf{r = \dfrac{1}{10}}$

$:\implies \bf{r = 0.1}$

$\boxed{\therefore \bf{r = 0.1}}$

Hence the distance between the two charges is 0.1 m.