Question

calculate the force between two electric charge and 5uc and -2uc seperated by a distance of 10 cm (1/4 r0=9×10^9Nm2/c2 ​

Answer 1

Answer :-

Given :-

• $\sf q_1 = 5 \mu C$
• $\sf q_2 = -2 \mu C$
• $\sf r = 10 cm$

To Find :-

• Force between two electric charge

Solution :-

Converting the units into SI system :-

• $\sf q_1 = 5 \mu C = 5 \times 10^{-6} C$
• $\sf q_2 = 2 \mu C = 2 \times 10^{-6} C \: ( \: only \: magnitude \: )$
• $\sf r = 10 cm = 0.1 m$

Substituting the value in formula and solving :-

$\implies\sf F = \dfrac{kq_1q_2}{r^2}$

$\implies\sf F = \dfrac{ 9 \times 10^9 \times 5 \times 10^{-6} \times 2 \times 10^{-6} }{(0.1)^2}$

$\implies\sf F = \dfrac{90 \times 10^{-6-6+9}}{1 \times 10^{-2}}$

$\implies\sf F = \dfrac{90 \times 10^{-3}}{10^{-2}}$

$\implies\sf F = 90 \times 10^{-3+2}$

$\implies\sf F = 90 \times 10^{-1}$

$\implies\sf F = 9$

Force of attraction = 9 N

Answer 2

Given :-

• ${\sf{q_{r}}}$ = $5μ$C
• ${\sf{q_{r}}}$ = $-2μ$C
• r = $10$cm

To find :-

• Force between two electric charge.

Solution :-

★ Converting the units into SI system :

• ${\sf{q_{r}}}$=$5μ$C = ${\sf{5~×~{10}^{ - 6}}}$C
• ${\sf{q_{r}}}$=$2μ$C = $2~×~{10}^{ - 6}$C${\sf{(only \: magnitude)}}$
• r = $10$cm =$0.1$m

★ Substituting the value :

$\large\dag$ Formula :

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ Solving :

:$\implies$F = $\large{\bf{\frac{k q_{1} q_{2} }{ {r}^{2} }}}$

$~~~~~$:$\implies$F = $\large{\bf{\frac{9 \: \times \: {10}^{9} \: \times \: 5 \: \times \: {10}^{ - 6} \: \times \: 2 \: \times \: {10}^{ - 6} }{(0.1 {)}^{2} }}}$

$~~~~~~~~~~$:$\implies$F = $\large{\bf{\frac{90 \: \times \: 1 {0}^{ - \: 6 \: - \: 6 \: + \: 9} }{1 \: \times \: 1 {0 }^{ - 2} }}}$

$~~~~~~~~~~~~~~~$:$\implies$F = $\large{\bf{ \frac{90 \: \times \: 1 {0}^{ - 3} }{1 {0}^{ - 2} } }}$

$~~~~~~~~~~~~~~~~~~~~$:$\implies$F = ${\bf{90 \: \times \: 1 {0}^{ - 3 \: + \: 2}}}$

$~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$F = ${\bf{90 \: \times \: 1 {0}^{ - 1} }}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$$\large{\underline{\boxed{\pink{\bf{F~=~9}}}}}$★

$\large\dag$ Hence,

• Force of attraction = $\underline{\bf{9~N}}$$\large{\bf\green{✓}}$