four equal charges of each +2mc are placed at the corner of side 50 cm.find the net force on any of the charges.
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Answer:
each are placed at the corners of sides 50cm. What are the force on any of the charges?
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Let each charge be q. We denote side length of the cube by d.Starting from left lower corner of the cube naming it A and moving anti clockwise we name the successive corners byB,and D respectively.We take coordinate system such that x-axis is along AB side and y-axis is along AD side.i and j unit vectors along x and y axis respectively. Now,we consider forces on charge atB.
Force due to charge atA, F(BA)=(kq^2/d^2)i
Force due to change atC, F(BC)=-(kq^2/d^2)j
Forc due to charge at D, F(BD)=kq^2/2d^2x(cos (pi/4)i- cos(pi/4))=kq^2/d^2(1/2root2i-1/2root2j)
Adding these three forces vectorially,we have,
F=kq^2/d^2(1+ 1/2root2)i-kq^2/d^2(1+1/2root2)j=kq^2/d^2(1+1/2root2)(I-j).
Now,take, q=2x10^-6C, d=0.5m and k=9x10^9 Nm^2/C^2. Find, with these values Fx andFy that is x and y components of F and then
|F|=under root of(Fx^2+Fy^2). I hope you will get numerical value by your self.