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Grade 12Electrostatics

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

Profile image of pooja  agrawal
13 Years agoGrade 12
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5 Answers

Profile image of atul shukla
13 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

Profile image of anurag singh
13 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

Profile image of Rohan Kumar
13 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

Profile image of ASh
8 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

Profile image of Rishi Sharma
6 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.


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