Question Number 46374 by Umar last updated on 24/Oct/18
$${At}\:{each}\:{of}\:{three}\:{corners}\:{of}\:{a}\:{square}\: \\ $$$${of}\:{side}\:\mathrm{2}{cm}\:{is}\:{placed}\:{a}\:{point}\:{charge}\: \\ $$$${of}\:{magnitude}\:\mathrm{3}\mu{C}. \\ $$$${What}\:{will}\:{be}\:{the}\:{magnitude}\:{and}\: \\ $$$${direction}\:{of}\:{the}\:{resultant}\:{force}\:{on}\:{a} \\ $$$${point}\:{charge}\:−\mathrm{1}\mu{C}\:{if}\:{it}\:{were}\:{placed} \\ $$$$\:\:\:\left({a}\right)\:\:{at}\:{the}\:{centre}\:{of}\:{the}\:{square}? \\ $$$$\:\:\:\left({b}\right)\:\:{at}\:{the}\:{vacant}\:{corner}\:{of}\:{the}\:{sqre}? \\ $$
Commented by Umar last updated on 24/Oct/18
$${please}\:{help} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Oct/18
$${let}\:{Square}\:{ABCD}... \\ $$$${at}\:\:{A}\left(\mathrm{0},\mathrm{0}\right)\:{charge}\:\mathrm{3}\mu{C} \\ $$$${at}\:{B}\left(\mathrm{2},\mathrm{0}\right)\:{charge}\:\mathrm{3}\mu{C} \\ $$$${at}\:{C}\left(\mathrm{2},\mathrm{2}\right)\:{no}\:{charge} \\ $$$${at}\:{D}\left(\mathrm{0},\mathrm{2}\right)\:{charge}\:\mathrm{3}\mu{C} \\ $$$${at}\:{E}\left(\mathrm{1},\mathrm{1}\right)\:{the}\:{centre}\:{charge}\:\left(−\mathrm{1}\mu{C}\right) \\ $$$${force}\:{on}\:{E}\:{by}\:{A}\:{is}\:{F}_{{EA}} {attractive} \\ $$$${force}\:{on}\:{E}\:{by}\:{B}\:\:{is}\:{F}_{{EB}\:\:{attractive}} \\ $$$${force}\:{on}\:{E}\:{by}\:{D}\:{is}\:{F}_{{ED}\:{attractive}} \\ $$$${but}\:{force}\:{F}_{{EB}} \:{is}\:{ewual}\:{in}\:{magnetude}\:{with} \\ $$$${F}_{{ED}\:} \:{but}\:{opposite}\:{in}\:{direction}\:{so}\:{cancell}\:{each}\:{other} \\ $$$${so}\:{net}\:{force}\:{on}\:{E}\:{is}\:{F}_{{EA}} \\ $$$$\overset{\rightarrow} {{F}}_{{EA}} =\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\frac{\left(\mathrm{1}×\mathrm{10}^{−\mathrm{6}} \right)\left(\mathrm{3}×\mathrm{10}^{−\mathrm{6}} \right)}{\left(\sqrt{\mathrm{2}}\:×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} }\:\:{direction}\:{E}\rightarrow{A} \\ $$$$=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{3}×\mathrm{10}^{−\mathrm{12}} }{\mathrm{2}×\mathrm{10}^{−\mathrm{4}} }=\mathrm{135}{N} \\ $$$${second}\:{question}... \\ $$$${when}\:{charge}\left(−\mathrm{1}\mu{C}\right)\:{placed}\:{at}\:{corner}\:{C} \\ $$$${force}\:{are}\:{F}_{{CB}} \:\:,\:{F}_{{CD}\:} \:\:{and}\:{F}_{{CA}} \\ $$$${F}_{{CB}} \:\:{magnetude}={F}_{{CD}} \:\:{magnetude}\:{say}\left({F}\right) \\ $$$${F}=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\left(\mathrm{1}×\mathrm{10}^{−\mathrm{6}} \right)\left(\mathrm{3}×\mathrm{10}^{−\mathrm{6}} \right)}{\left(\mathrm{2}×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} }{N}=\mathrm{6}.\mathrm{75}{N} \\ $$$${Resultant}\:{force}\:{of}\:{F}_{{CB}} \:\:{and}\:{F}_{{CD}} ={F}_{{R}} \left({say}\right) \\ $$$${F}_{{R}} =\sqrt{{F}^{\mathrm{2}} +{F}^{\mathrm{2}} +\mathrm{2}{F}×{Fcos}\mathrm{90}^{\mathrm{0}} }\:=\sqrt{\mathrm{2}}\:{F}\:\:{direction}\:{C}\rightarrow{A} \\ $$$${F}_{{R}} =\sqrt{\mathrm{2}}\:×\mathrm{6}.\mathrm{75}\:{N}=\mathrm{1}.\mathrm{41}×\mathrm{6}.\mathrm{75}\approx\mathrm{9}.\mathrm{52}{N} \\ $$$${now}\:{F}_{{CA}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\left(\mathrm{1}×\mathrm{10}^{−\mathrm{6}} \right)\left(\mathrm{3}×\mathrm{10}^{−\mathrm{6}} \right)}{\left(\mathrm{2}\sqrt{\mathrm{2}}\:×\mathrm{10}^{−\mathrm{2}} \right)^{\mathrm{2}} }\:{direction}\:{C}\rightarrow{A} \\ $$$${F}_{{CA}} =\frac{\mathrm{27}}{\mathrm{8}}×\mathrm{10}=\mathrm{33}.\mathrm{75}{N} \\ $$$${so}\:{net}\:{force}=\left(\mathrm{9}.\mathrm{52}+\mathrm{33}.\mathrm{75}\right){N}=\mathrm{43}.\mathrm{27}\:{N}\:{direction}\: \\ $$$${C}\rightarrow{A} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$
Commented by Umar last updated on 31/Oct/18
$${thank}\:{you}\:{sir} \\ $$