Two students studied electric interactions using the charged beads and a spring.

Question

Two students studied electric interactions using the charged beads and a spring. They attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge q. This stretches the spring by 3.00mm. Then they gave the beads the opposite charges (+q and –q). This compressed the spring by 5.00mm. During experimentation, they also measured length of unstretched spring and charge q but these data were lost. Please help to restore the length of the unstretched spring using the above data. What relation between q and spring constant can be extracted from these data?

Later, it was found that the spring constant is equal to 2.50 N/m. What was the magnitude of the charge on each of the beads?  

Explanation / Answer

let spring constant be k1.

let length of the spring be L mm.

then when charges are of same sign, distance between them=(L+3) mm

force=k*q*q/((L+3)*0.001)^2

where k is coloumb;s constant=9*10^9

this force is equal to spring force=spring constant*elongation

==>k*q^2/((L+3)*0.001)^2=k1*3*0.001

==>k*q^2/k1=((L+3)*0.001)^2*0.003...(1)

when charges are of opposite sign, distance between them=L-5 mm

equating electrical force with spring force,

k*q^2/((L-5)*0.001)^2=k1*0.005

==>k*q^2/k1=((L-5)*0.001)^2*0.005...(2)

from 1 and 2,

(L+3)^2*3=(L-5)^2*5

==>3*(L^2+6*L+9)=5*(L^2-10*L+25)

==>2*L^2-68*L+98=0

==>L=32.492 mm

so original length of the spring is 32.492 mm

part b:

from equation 1,

k*q^2/k1=((L+3)*0.001)^2*0.003

using k=9*10^9,k1=2.5, L=32.492

we get q^2=1.04973*10^(-15)

==>q=3.24*10^(-8) C

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