Question

Solve the one in attachment:​

Answer 1

$\huge\tt{\red{\underline{Given:}}}$

★Force = 20N (F)

★ Distance = 30cm =0.3m (r)

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★value of charges of both bodies

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We will use Columb's Law

$\huge\tt{\red{\underline{Answer:}}}$

We have,

Force = 20N

Distance between the charged particles = 0.3m

Now, let the charge on first body be $q_{1}$ and on second body be $q_{2}$

_______________________________________

Atq,

${\underline{\boxed{\red{q_{1}+q_{2}=30\mu C}}}}$ .......... (1)

Now by Columb's Law,

$\large\green{\boxed{F=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}q_{2}}{r^{2}}}}$

where

• F is the force
• $q_{1}$&$q_{2}$ are the two charges
• r is the distance of seperation.

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

$\implies 20N =\dfrac{q_{1}q_{2}\times 9\times 10^{9}}{(0.3m)^{2}}$

$\implies 20N = \dfrac{q_{1}q_{2}\times 9\times 10^{9}}{(0.09m^{2}}$

$\implies 20N = \dfrac{q_{1}q_{2}\times 9\times 10^{9}\times 100}{(9m^{2})}$

$\implies 20N = \dfrac{q_{1}q_{2}\times \cancel{9}\times 10^{9}\times 100}{(\cancel{9}m^{2})}$

$\implies 20N =q_{1}q_{2}\times10^{11}$

$\implies q_{1}q_{2}=\dfrac{20C^{2}}{10^{11}}$

${\underline{\boxed{.°. q_{1}q_{2}=20\times10^{-11}C^{2}}}}$

Now, we should find $q_{1}-q_{2}$in order to find $q_{1}$&$q_{2}$.

______________________________________

So we can find $q_{1}-q_{2}$ as

$\large{\boxed{\red{(q_{1}-q_{2})^{2}=(q_{1}+q_{2}^{2})-4q_{1}q_{2}}}}$

$\implies (q_{1}-q_{2})^{2}=(30\mu C) ^{2}-4\times 20\times 10^{-11}C^{2}$

$\implies (q_{1}-q_{2})^{2}=900\mu^{2}C^{2}-80\times10^{-11}C^{2}$

$\implies (q_{1}-q_{2})^{2}=900\mu^{2}C^{2}-800\times(10^{-6}) ^{2} C^{2}$

$\implies (q_{1}-q_{2})^{2}= 900\mu^{2}C^{2}-800\mu^{2}C^{2}$

$\implies (q_{1}-q_{2})^{2}= 100\mu^{2}C^{2}$

$\implies (q_{1}-q_{2})^{2}= (10\mu C) ^{2}$

${\underline{\boxed{\red{.°. q_{1}-q_{2}=10\mu C}}}}$. ............ (2)

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On adding equations 1&2 ,

$\implies q_{1}+q_{2}+q_{1}-q_{2}=30\mu C+10\mu C$

$\implies q_{1}+\cancel{q_{2}}+q_{1}-\cancel{q_{2}}=40\mu C$

$\implies 2q_{1}=40\mu C$

$\implies q_{1}=\dfrac{40\mu C}{2}$

${\underline{\boxed{\purple{.°.q_{1}=20\mu C}}}}$

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Now, from 1,

$\implies q_{1}+q_{2}=30\mu C$

$\implies 10\mu C+q_{2}=30\mu C$

$\implies q_{2}=(30-20) \mu C$

${\underline{\boxed{\orange{.°. q_{2}=10\mu C}}}}$

Therefore the two charges are $10\mu C$ & $20\mu C$.

Answer 2

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

$\green{\tt{\therefore{q_{1}=20\:\mu c}}}$

$\green{\tt{\therefore{q_{2}=10\:\mu c}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Sum \: of \: two \: charges = 30 \: \mu c \\ \\ \tt: \implies Force(F) = 20 \: N \\ \\ \tt: \implies distance \: between \: them(r) = 0.3 \: m \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies q_{1} = ? \\ \\ \tt: \implies q_{2} = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies q_{1} + q_{2} = 30 \: \mu c \\ \\ \tt: \implies q_{1} + q_{2} = 3 \times {10}^{ - 5} \: c-----(1) \\ \\ \bold{As \: per \: Coulomb's \: law} \\ \tt: \implies |F|= \dfrac{1}{4\pi \epsilon _{o} } \dfrac{ q_{1} q_{2} }{ {r}^{2} } \\ \\ \tt: \implies 20 = 9 \times {10}^{9} \times \dfrac{ q_{1} q_{2} }{ {0.3}^{2} } \\ \\ \tt: \implies 20 = {10}^{11} \times q_{1} q_{2} \\ \\ \tt: \implies q_{1} q_{2} = 2 \times {10}^{ - 10} \: {c} - - - - - (2) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies (q_{1} - q_{2})^{2}= { (q_{1} + q_{2}})^{2} - 4 q_{1} q_{2} \\ \\ \tt: \implies (q_{1} - q_{2})^{2} = {(3 \times 10^{ - 5}) }^{2} - 4 \times 2 \times {10}^{ - 10} \\ \\ \tt: \implies (q_{1} - q_{2})^{2} =9 \times 10^{ - 10} - 8 \times 10^{ - 10} \\ \\ \tt: \implies (q_{1} - q_{2})^{2}= 1 \times 10 ^{ - 10} \\ \\ \tt:\implies q_{1}-q_{2}=\sqrt{10^{-10}}\\\\ \tt:\implies q_{1}-q_{2}=10^{-5}\:c-----(3)\\ \\ \text{From \: (1) \: and \: (3)} \\ \\ \green{ \tt: \implies q_{1} = 20 \: \mu c} \\ \\ \green{ \tt: \implies q_{2} = 10\: \mu c}$