Question

four-point charges are located at the corners of a square, 1.00 m by 1.00 m. On each
of two diagonally opposite corners are 1.00 μC charges. On each of the other two
corners are -1.00 μC charges. What is the magnitude of the force on one of the
positive charges?

Answer 1

Given:-

$\\charge(q)=1.0\mu C\\[5pt] distance(a)=1.0m\\[5pt] Find \,magnitude\, of\, the \,force \,on\, one\, of\, the\,positive\, charges$

Now,

$\\Along \,the\,diagonals\, the \,centre\, of\, the\, square .\\[5pt] The\, charges\, attract\, each\, other\, with\, a \,forcek\, =\frac{q^{2}}{a^{2}}\\[5pt]$

$\therefore \\ Their\;Resultant\,(R)=2Fcos45=(1.414\,k.\frac{q^2}{a^2})\\[5pt] Also\,The\, diagononally\,opposite\,charges\, repel \,each\,other\,\\[5pt] with\, a\, force\,(F')=k.(\frac{q^2}{2a^2})\, acting\, along\, the\,diagonal\,outward.\\[5pt]$

$\\so,\\[5pt] the\,net\,force\,is \,the\,\sum F=\left ( 1.414\,k.\frac{q^2}{a^2}-0.5\,k.\frac{q^2}{a^2} \right )\\[10pt] Now\,\\[5pt] Force (F)=\left ( 0.914\,k.\frac{q^2}{a^2} \right )\\[10pt] Magnitude\,of\,force\,F=\left ( 0.914\,\times (9\times 10^{9}N\frac{m^2}{c^2})\times \frac{(1\times 10^{-6}C)^2}{(1.0m)^2} \right )\\[10pt] Magnitude\,of\,force\,F=0.0081\,N\\[10pt] Along\,the\,diagonal\,towards\,the\,centre\,of\,the\,square\,$

Answer:-

The magnitude of the force on one of the positive charges

Force(F)=0.0081 N