Question

${\mathfrac{\huge{\red{QUESTION}}}}$


${\texttt{\blue{Sub :- Physics}}}$


${\texttt{\green{Topic :- Electrical charge}}}$


${\texttt{Two positively fixed point charges}}$
${\texttt{which are separated by a}}$
${\texttt{distance of 0.1 m repeal each }}$
${\texttt{other with a force of 18N,}}$
${\texttt{if the sum of charges = 9 C .}}$
${\texttt{Then find the magnitude of both}}$
${\texttt{charges .}}$



${\matcal{Thanks}}$


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Answer 1

$\pink{ \bf{Heya !! }}$

$\purple{ \bf{Here \: is \: the \: answer !! }}$

$\red {\texttt{The information given are : }}$

♦The distance of separation between the positively charges point charges is 0.1 m

♦ The force of repulsion is 18N

♦The sum of charges is 9 μC

$\blue{ \texttt{We need to find the }}$
$\blue{ \texttt{ magnitude of the charges ! }}$

According to Coulomb's Inverse square law ,
The attractive force or repulsive force between two point charges is proportional to the product of magnitude of the charges and inversely proportional to the square of the distance between them.

$F \: = k \frac{q_1q_2}{ {r}^{2} }$

where, k = proportionality constant

$k = 9 \times {10}^{9} N {m}^{2} {C}^{ - 2}$

Substituting the given values in the formula , we get
$18 = 9 \times {10}^{9} ( \frac{q_1q_2}{1 \times {10}^{ - 2} } )$

$q_1q_2 = \frac{8 \times {10}^{ - 2} }{9 \times {10}^{9} }$

$q_1q_2 = 2 \times {10}^{ - 11}$
For our convenience , we write it as

$q_1q_2 = 20 \times {10 }^{ - 12}$
$q_1q_2 = 20 \: {μC}^{2}$

We know that sum of the charges is 9

$q_1 + q_2 = 9$
$q_2 = 9 - q_1$

Then,
$q_1(9 - q_1) = 20 \\ {q_1}^{2} - 9q_1 + 20 = 0$
On solving this , we get
${q_1}^{2} - 5q_1 - 4q_1 + 20 = 0 \\ q_1(q_1 - 5) \: - 4(q_1 - 5) =0$
$(q_1 - 5) = 0 \\ (q_1 - 4) = 0$
Then
$\: q_1 \: = 4μC \: or \: 5μC$
Therefore ,
$if \: q_1 = 4μC, \: then \: q_2 = 5μC \\ if \: q_1 = 5μC ,\: then \: q_2 = 4μC$

$\orange{ \large{ \texttt{Hope This Helps You :) }}}$

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