# an infinite number of charges each equal to q are placed along x axis at x=1, x=2,x=4,x=8 and so on find the potential and electric field at the point x=0 due to this set of charges

atul shukla
31 Points
10 years ago

potential at x=0

=kq(1+(1/2)+(1/4)+(1/8)..........up to infinite).... it will become GP series

(1+(1/2)+(1/4)+(1/8)..........up to infinite)=2 (sum toinfinity = a / (1 -r )) a=1 and r=1/2

so ans is 2kq

anurag singh
19 Points
10 years ago

potential =kq/r in terms of magnitude

in this question the net potential is sum of all potentials

kq+kq/2+kq/4.........infinity

it is an infinite gp

net potential=a/1-r=kq/1-1/2

2kq

similarly magnitude of net field=summation of kq/r2

kq+kq/4+kq/16......infinity

=kq/1-1/4

=4/3kq

Rohan Kumar
18 Points
10 years ago

Suppose that charges q1,q2,q3,... are placed at distances r1,r2,r3,... from the origin. Then,electric feild at the origin due to the system of charges,

E=1/4∏ε{(q1/r12)+(q2/r22)+(q3/r32)+.....}

Here q1+q2+q3+...=q

and r1=1,r2=2,r3=4,...

Therefore,

E=1/4∏ε{(q/12)+(q/22)+(q/42)+.....}

E=1/4∏ε.q{(1/12)+(1/22)+(1/42)+.....}

E=1/4∏ε.q.S∞(S infinity)

where, S (infinity)=a/1-r=1/1-1/4=4/3

Hence,

E=1/4∏ε.q*4/3

E=q/3∏ε

ASh
62 Points
5 years ago
Here q1+q2+q3+...=q

and r1=1,r2=2,r3=4,...

Therefore,

E=1/4∏ε{(q/12)+(q/22)+(q/42)+.....}

E=1/4∏ε.q{(1/12)+(1/22)+(1/42)+.....}

E=1/4∏ε.q.S∞(S infinity)

where, S (infinity)=a/1-r=1/1-1/4=4/3

Hence,

E=1/4∏ε.q*4/3

E=q/3∏ε

Rishi Sharma