Question

Two positive charges distant 0.1 metre apart repel each other with a force of 18 newton is the sum of the charges be 9 microcoulomb then calculate their separate values

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{q_{1}=5\:\mu c}}}$

$\green{\tt{\therefore{q_{2}=4\:\mu c}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Sum \: of \: two \: charges = 9 \: \mu c \\ \\ \tt: \implies Force(F) = 18 \: N \\ \\ \tt: \implies distance \: between \: them(r) = 0.1 \: m \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies q_{1} = ? \\ \\ \tt: \implies q_{2} = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies q_{1} + q_{2} = 9 \: \mu c \\ \\ \tt: \implies q_{1} + q_{2} = 9 \times {10}^{ - 6} \: c \\ \\ \bold{As \: per \: Coulomb's \: law} \\ \tt: \implies |F|= \frac{1}{4\pi \epsilon _{o} } \frac{ q_{1} q_{2} }{ {r}^{2} } \\ \\ \tt: \implies 18 = 9 \times {10}^{9} \times \frac{ q_{1} q_{2} }{ {0.1}^{2} } \\ \\ \tt: \implies 18 = 9 \times {10}^{11} \times q_{1} q_{2} \\ \\ \tt: \implies q_{1} q_{2} = 20 \times {10}^{ - 12} \: {c} - - - - - (2) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies (q_{1} - q_{2})^{2}= { (q_{1} + q_{2}})^{2} - 4 q_{1} q_{2} \\ \\ \tt: \implies (q_{1} - q_{2})^{2} = {(9 \times 10^{ - 6}) }^{2} - 4 \times 20 \times {10}^{ - 12} \\ \\ \tt: \implies (q_{1} - q_{2} )^{2}=81 \times 10^{ - 12} - 80 \times 10^{ - 12} \\ \\ \tt: \implies q_{1} - q_{2} = \sqrt{1 \times 10 ^{ - 12}} \\\\\ \tt:\implies q_{1}-q_{2}=10^{-6}\:c - - - - - (3) \\ \\ \text{From \: (1) \: and \: (3)} \\ \\ \green{ \tt: \implies q_{1} = 5 \: \mu c} \\ \\ \green{ \tt: \implies q_{2} = 4 \: \mu c}$

Answer 2

....

$\tt{\huge{\purple{ .................. }}}$

QUESTION -

Two positive charges distant 0.1 metre apart repel each other with a force of 18 newtons.

The sum of the charges be 9 microcoulombs.

Then calculate their separate values.

SOLUTION -

In the above Question, the following information is given...

Two positive charges distant 0.1 metre apart repel each other with a force of 18 newton .

The sum of the charges is 9 microcoulombs.

So,

We know that :

$1 \: micro \: Coulumb = 1 \times { 10} ^ { -6} \: C$

$\leadsto q_{1} + q _ { 2 } = 9 \times { 10} ^ { -6 } \: C ............. ( 1 )$

$F = \dfrac{ k q_{1} q_{2} }{ { r } ^ 2 } = 18 \: N$

$Where \: k = 9 \times { 10} ^ { -9 }$

$Solving \: we \: get \::-$

$q_{1} \times q_{2} = 12 \times {10} ^ { -20 } \: C$

We know that :

$( a - b ) ^ 2 = ( a + b ) ^ 2 - 4ab$

Using this identity we Obtain :

$q_{1} - q_{2} = 1 \times { 10} ^ { -12 } \: C ................. ( 2 )$

Adding equations 1 and 2, we get :

$=> 2q_{1} = 1 \times {10} ^ { -5 } C \\ \\ => q_{1} = 5 \times {10} ^ { -6 } C = 5 { \nu} C$

Similarly solving we get :

$=> q_{2} = 4 { \nu } C$