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This is second of our many quizzes on Coulomb's law. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Coulomb's law Quiz 2

physics

 In case $(1)$ the charge $+Q$ uniformly spread over a semi-circular ring and a charge $+Q$ placed at center of semi-circle $P$. In the Second case $(2)$ two equal positive charges $+Q$ are separated by distance $R$. The magnitude of force. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/51bd0fce-68bf-4ec8-86f7-f5b309e6648f7805507125792810830.png"/> 

Due to presence of charge, force experience by another charge is $F=\dfrac{K{{q}_{1}}{{q}_{2}}}{r}$. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/33023ccf-4ed7-458f-8ea8-fd92565545ec1545860373432817336.png"/>$ \\ $ Small charge $dQ$ for small arc of angle $dQ$ and length $dx$ is $ \\ dQ=\dfrac{Q}{\pi R}dx \\ \Rightarrow dQ=\dfrac{Q}{\pi R}Rd\theta  \\ \Rightarrow dQ=\dfrac{Qd\theta }{\pi } \\ $ Small amount of force due to this small charge is $ \\ dF=\dfrac{KQdQ}{{{R}^{2}}} \\ \Rightarrow dF=\dfrac{K{{Q}^{2}}d\theta }{\pi {{R}^{2}}} \\ $ As the $\cos\theta$ components will get cancelled and only sin $\theta$ components add up. $ \\ {{F}_{net}}=\int{dF\sin \theta } \\ \Rightarrow {{F}_{net}}=\int\limits_{0}^{\pi }{\dfrac{K{{Q}^{2}}}{\pi {{R}^{2}}}}\sin \theta d\theta  \\ \Rightarrow \dfrac{K{{Q}^{2}}}{\pi {{R}^{2}}}\left[ -\cos \theta  \right]_{0}^{\pi } \\ \Rightarrow {{F}_{net}}=\dfrac{K{{Q}^{2}}}{\pi {{R}^{2}}} \\ $ For arrangement $ \left( 2 \right)$ $\\{{F}_{2}}=\dfrac{KQQ}{{{R}^{2}}} \\ \Rightarrow {{F}_{2}}= \dfrac{K{{Q}^{2}}}{{{R}^{2}}} \\ $ Thus force in $\left( 1 \right)$ is less than in $\left( 2 \right).$

 The magnitude of electric force at point $O$ is $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cd2ed22c-6ef5-451d-9786-f6819d08dcc17850355563960104387.png"/>

Small force at $O$ due to small charge element $dQ$ is $dF=\dfrac{K2QdQ}{{{x}^{2}}}$ where $x$ is distance between point $O$ and charge element $dQ$. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/73eedc20-cd0d-4603-bc6a-4e75b7b309369196771040986596684.png"/>$ \\ $ Let the small amount of force due to a small charge element dQ be dF. $ \\ $ The Vertical components of dF from corresponding upper and lower parts of  the rod i.e., ${dFsin}\theta$ will cancel out. $ \\ $ Only the ${dFcos}\theta$ components contribute to the total force. $ \\ $ Thus, total force $F=\int{2dF\cos \theta}$ will always be smaller than $F'=\dfrac{K2{{Q}^{2}}}{{{P}^{2}}}$

A square of side $a$ has charges $+Q$ on three corners $A,B$ and $D$. A new charge $+Q$ is placed on point $P$ as shown. Magnitude of electric force at $P$ is $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4727bf27-8435-4b65-80b3-6178c838a6ca254451004776457646.png"/> 

Use principle of superposition at point $P$ and use formula $\dfrac{K{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4e2ad70f-be42-4c19-9a0b-6d26be85d1ca1031668766664562039.png"/> $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/04629838-d68e-4b51-aee0-e98484ffa58c3989345801609982225.png"/>$ \\ \text{In figure }\left( 1 \right) \\DP=AP=\sqrt{{{\left( \dfrac{a}{2} \right)}^{2}}+{{a}^{2}}}=\sqrt{\dfrac{5{{a}^{2}}}{2}}=a\sqrt{\dfrac{5}{2}} \\ \Rightarrow {{F}_{PA}}={{F}_{PD}}=\dfrac{K{{Q}^{2}}}{{{\left( \dfrac{a\sqrt{5}}{2} \right)}^{2}}} \\ \Rightarrow {{F}_{PA}}=\dfrac{4K{{Q}^{2}}}{5{{a}^{2}}} \\ $ As the sine components cancel out their resultant is $  =2F\cos \theta  \\ \Rightarrow 2\left( \dfrac{4K{{Q}^{2}}}{5{{a}^{2}}} \right)\sqrt{\dfrac{2}{5}} \\ \text{By figure }\left( 2 \right)\text{resultant of } \\ 2F\cos \theta \text{ and }+{{{\vec{F}}}_{PB}} \\ \Rightarrow {{F}_{eq}}=\sqrt{{{\left( 2F\cos \theta  \right)}^{2}}+{{F}_{PB}}^{2}} \\ \Rightarrow {{F}_{eq}} =\sqrt{\left( \dfrac{8k{{Q}^{2}}}{5{{a}^{2}}} \right)\left( \dfrac{2}{5} \right)+\dfrac{4K{{Q}^{2}}}{{{a}^{2}}}} \\ \Rightarrow {{F}_{eq}} =\dfrac{4K{{Q}^{2}}}{{{a}^{2}}}\sqrt{\dfrac{29}{25}} \\ \therefore {{F}_{eq}} =\dfrac{4}{5}\dfrac{K{{Q}^{2}}}{{{a}^{2}}}\sqrt{29}$

 If a unit positive charge placed at center of following then<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d02c805d-e75d-4544-af83-70cdd33b2e87648068309173697727.png"/>

See the geometry and determine if the system is symmetrical around point $O$.For a symmetrical distribution, the net force becomes zero. 

 &nbsp;$\text{Symmetrical distribution of equal charges} \\ \text{around a point always has a total equivalent force=0 N,} \\ \text{as the charges and the distance involved is equal}\text{.} \\ \text{Due to this, the forces being opposite, the equivalent becomes zero}\text{.} \\ \text{In Case }\left( \text{1} \right) \\ \text{Due to symmetry, force at center is zero} \\ \text{In Case }\left( \text{2} \right) \\ \text{Force at symmetry triangle is zero}\text{.} \\ \text{In case }\left( \text{3} \right)\,\text{Force at center of square is also zero due to symmetry}\text{.}$

In $X-Y$ plane two charges are placed at ${{\vec{r}}_{1}}$ and ${{\vec{r}}_{2}}$ respectively the magnitude of force on one by other is<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/52b5cb37-2874-404c-b478-24c6c6cfacc52237781968799102314.png"/> 

From Coulomb’s law, $F=\dfrac{K{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$, $r$ is distance between two charge particle. 

 ${{{\vec{r}}}_{1}}=2\hat{i}+3\hat{j},{{{\vec{r}}}_{2}}=6\hat{i}+6\hat{j} \\ \Rightarrow {{{\vec{r}}}_{2}}-{{{\vec{r}}}_{1}}=4\hat{i}+3\hat{j} \\ \Rightarrow \left| {\vec{r}} \right|=\sqrt{{{4}^{2}}+{{3}^{2}}} \\ \Rightarrow \left| {\vec{r}} \right|=\text{5 meter} \\ \text{As the force between two point charges is} \\ F=k\dfrac{Q\times Q}{{{\left| {\vec{r}} \right|}^{2}}} \\ \therefore F=\dfrac{k{{Q}^{2}}}{{{5}^{2}}}=\dfrac{k{{Q}^{2}}}{25}$