Question

Find the force? Explain in detail

Answer 1

Answer:

where is force

force is always around us

Answer 2

Here,

• $\sf{q_A=q_B=q_C=q_D=3\times10^{-6}\ C}$

• $\sf{r_{AB}=r_{BC}=r_{CD}=r_{DA}=0.1\ m}$

• $\sf{r_{AC}=r_{BD}=0.1\sqrt2\ m}$

The force acting at B due to A,

$\longrightarrow\bf{F_{BA}}=\sf{\dfrac{k\,q_A\,q_B}{(r_{AB})^2}}\,\bf{\hat i}$

$\longrightarrow\bf{F_{BA}}=\sf{\dfrac{9\times10^9\times3\times10^{-6}\times3\times10^{-6}}{(0.1)^2}}\,\bf{\hat i}$

$\longrightarrow\bf{F_{BA}}=\sf{8.1}\,\bf{\hat i}$

The force acting at B due to C,

$\longrightarrow\bf{F_{BC}}=\sf{\dfrac{k\,q_B\,q_C}{(r_{BC})^2}}\,\bf{\hat j}$

$\longrightarrow\bf{F_{BC}}=\sf{\dfrac{9\times10^9\times3\times10^{-6}\times3\times10^{-6}}{(0.1)^2}}\,\bf{\hat j}$

$\longrightarrow\bf{F_{BC}}=\sf{8.1}\ \bf{\hat j}$

The force acting at B due to D,

$\longrightarrow\bf{F_{BD}}=\sf{\dfrac{k\,q_B\,q_D}{(r_{BD})^2}}\ \left(\sf{\cos45^o}\,\bf{\hat i}+\sf{\sin45^o}\,\bf{\^j}\right)$

$\longrightarrow\bf{F_{BC}}=\sf{\dfrac{9\times10^9\times3\times10^{-6}\times3\times10^{-6}}{\left(0.1\sqrt2\right)^2\sqrt2}}\,\left(\bf{\hat i+\hat j}\right)$

$\longrightarrow\bf{F_{BC}}=\sf{\dfrac{8.1}{2\sqrt2}}\ \bf{\hat i}+\sf{\dfrac{8.1}{2\sqrt2}}\ \bf{\hat j}$

Hence net force at B is,

$\longrightarrow\bf{F_B=F_{BA}+F_{BC}+F_{BD}}$

$\longrightarrow\bf{F_B}=\sf{8.1}\ \bf{\hat i}+\sf{8.1}\ \bf{\hat j}+\sf{\dfrac{8.1}{2\sqrt2}}\ \bf{\hat i}+\sf{\dfrac{8.1}{2\sqrt2}}\ \bf{\hat j}$

$\longrightarrow\bf{F_B}=\sf{8.1}\left(1+\sf{\dfrac{1}{2\sqrt2}}\right)\,\left(\bf{\hat i}+\bf{\hat j}\right)$

Its magnitude is,

$\longrightarrow\sf{F_B}=\sf{8.1}\left(1+\sf{\dfrac{1}{2\sqrt2}}\right)\sf{\sqrt{1^2+1^2}\ N}$

$\longrightarrow\sf{F_B}=\sf{8.1\left(\sqrt2+\dfrac{1}{2}\right)\ N}$

Taking $\sf{\sqrt2=1.414,}$

$\longrightarrow\sf{F_B}=\sf{8.1\left(1.414+\dfrac{1}{2}\right)\ N}$

$\longrightarrow\underline{\underline{\sf{F_B}=\sf{15.5\ N}}}$

The force at C is the same as that at B. Therefore,

$\longrightarrow\underline{\underline{\sf{F_C}=\sf{15.5\ N}}}$