4 charges equal to -Q placed at the 4 corners of a square and a charge q at the centre ,if system is in equilibrium q is
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Here since the system is in equilibrium hence the net force on each charge q will be 0. We are considering 1 such charge. Here:
Force F = Kq2/a2 +Kq2/a2 + the forces due to diagonal element
First lets find he resultant of the 1st two forces:
R = √2(Kq2/a2)
Now force due to diagonal charge q :
F1 = Kq2/(a√2)2 where a√2 is the length of the diagonal
F1 = Kq2/2a2
Force F2 due to Q :
F2 = KqQ/(a√2/2)2
Net force on q hence = R +F1+F2 = 0
√2Kq2/a2 +Kq2/2a2 +2KqQ/a2 = 0
Cancelling Kq/a2 on both sides:
Q = -(4√2 +2)q/8
Force F = Kq2/a2 +Kq2/a2 + the forces due to diagonal element
First lets find he resultant of the 1st two forces:
R = √2(Kq2/a2)
Now force due to diagonal charge q :
F1 = Kq2/(a√2)2 where a√2 is the length of the diagonal
F1 = Kq2/2a2
Force F2 due to Q :
F2 = KqQ/(a√2/2)2
Net force on q hence = R +F1+F2 = 0
√2Kq2/a2 +Kq2/2a2 +2KqQ/a2 = 0
Cancelling Kq/a2 on both sides:
Q = -(4√2 +2)q/8
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thank you.
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