 # 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

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

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

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

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

3 years ago
Hello students,
The solution of the above problem is in the attached file.
I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams. 