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Question

An infinite number of identical charges, each +q coulomb, are placed along x-axis at x=1 m,3 m,9 m,... Calculate the electric field at the point x=0 due to these charges.

A
14πε09q10 NC1
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B
14πε09q5 NC1
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C
14πε09q8 NC1
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D
14πε09q7 NC1
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Solution

The correct option is C 14πε09q8 NC1
Electric field at origin will be resultant of all electric fields due to individual charges.

E0=E1+E2+E3+...

Substituting the values of the electric field due to each charge, we get

E0=kq12+kq32+kq92+...

E0=kq[112+132+192+.....]

E0=q4πε0[1+19+181+...]...(1)

As we know that sum of infinite terms of Geometric progression,

S=a(1r)

Where a=1 is the first term and

r is the ratio of two consecutive terms, r=ba=191=19

Now, substituting the values in (1),

E0=q4πε0[a(1r)]

E0=q4πε0⎢ ⎢ ⎢1119⎥ ⎥ ⎥

E0=98(q4πε0)

Hence, option (a) is correct answer.

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