Question

Two bodies are opposite charged with 1000 µC and 400 µC charge. Find the force between the two charges if the distance between them is 0.5m. with explanation and proper steps!

Answer 1

Answer:

14400 N or 1.44 $\times 10^4$ N

Explanation:

Force between two charges is given by: $\frac{1}{4\pi \epsilon} \frac{q_1 q_2}{r^2}$

Experimental value of $\frac{1}{4\pi \epsilon}$ is 9 x 10^(9) and $\mathrm{\mu=10^{-6}}$

Substitute values of the required things:

$f = \small{9 \times {10}^{9}} \times \frac{(1000 \times {10}^{ - 6} ) \times(400 \times {10}^{ - 6}) }{(0.5) {}^{2} } \\ \\ f = \small{9 \times {10}^{9} } \times \frac{1000 \times 400 \times {10 }^{ - 6 -6 } }{0.25} \\ \\ f = 9 \times {10}^{9} \times (1 \times 4 \times {10}^{3 + 2} ) \times \frac{ {10}^{ - 12} }{0.25} \\ \\ f = 144 \times {10}^{9 + 5 - 12} = 144 \times {10}^{2}$

F = 14400N or 1.44 $\times 10^4$ N

Answer 2

$\huge\mathcal{\underbrace{SOLUTION:-}}$

$\bigstar\:\rm{\red{Force\:=\:\dfrac{k\:q_1\:q_2\:}{r^2}\:}}$

✍️ Here,

• $\rm{q_1\:=\:1000\:\mu\:C}$

• $\rm{q_2\:=\:400\:\mu\:C}$

• k = $\rm{\dfrac{1}{4\pi\epsilon_{o}}\:=\:9\times{10^{9}}\:}$

• r = 0.5m

• $\rm{\underline{\green{\mu\:=\:micro\:=\:10^{-6}\:}}}$

✍️ $\rm{1000\:\mu{C}\:=\:1000\times{10^{-6}}\:C}$

✍️ $\rm{400\:\mu{C}\:=\:400\times{10^{-6}}\:C}$

$\rm{\implies\:F\:=\:\dfrac{(9\times{10^9})\:\times\:1000\times{10^{-6}}\:\times\:400\times{10^{-6}}\:}{(0.5)^2}\:}$

$\rm{\implies\:F\:=\:\dfrac{36}{25}\:\times{10^{16}}\:\times{10^{-12}}\:}$

$\rm{\purple{\implies\:F\:=\:1.44\times{10^4}\:N}}$

✍️ Therefore, the force between the two charges is ‘$\mathcal{\underline{\blue{1.44\times{10^{4}}\:N}}}$’ .