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HETAV PATEL

Grade 11,

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A charge +q is fixed at each of the points x = x. , x = 3x. , x = 5x. and so on to infinity on the x-axis and a charge -q is fixed at each of the points x = 2x. , x = 4x. , x = 6x. and so on to infinity. Here x. is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4*pie*epsilon.*r). Then the potential at the origin to the above system is:-

A charge +q is fixed at each of the points x = x. , x = 3x. , x = 5x. and so on to infinity on the x-axis and a charge -q is fixed at each of the points x = 2x. , x = 4x. , x = 6x. and so on to infinity. Here x. is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4*pie*epsilon.*r). Then the potential at the origin to the above system is:-

Grade:12

2 Answers

Sameer Baheti
29 Points
7 years ago
Since potential is a scalar, the potential due to all charges separately can be written at the origin and added to get the answer. The potential due to all charges when added gives kq/x -kq/2x +kq/3x -kq/4x ...and so on to infinity. This can also be written as kq.ln2/x. As the expansion of ln(1+x) is x-x.x/2+x.x.x/3-x.x.x.x/4+ and so on for x belonging to [-1,+1]
ankit singh
askIITians Faculty 614 Points
3 years ago

Potential at origin due to positive charge

V(+)=x0​Kq​+3x0​Kq​+5x0​Kq​......∞
Similarly,
V(−)=−[2x0​Kq​+4x0​Kq​+6x0​Kq​......∞]
Net potential 
Vnet​=x0​Kq​−2x0​Kq​+3x0​Kq​−4x0​Kq​......∞
⇒Vnet​=x0​Kq​[1−21​+31​−41​......∞]
⇒Vnet​=x0​Kq​ln2

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