Question

an infinite number of charges each equal to 1nC are placed along the x-axis at distance x=1 m ,x=3 m, x=9 m....etc,the charges are the alternately positive and negative. at the x=1m the charge is x=9 m , the charge is -1nC at x=3 m the charges at +1nC..etc. A charge of +1c is placed at the origin. calculate the resultant force experienced by the charge at the origin

Answer 1

Answer:81×10^-17

Explanation:

The positive charge repule the charge at origin and the negative charge attract them therefore

F={(q1)(q2)×9×10^-9}/(r)^2

F={(1×1×10^-9)×9×10^-9}{(1/1^2)-(1/3^2)-(1/9^2)-(1/(3×9)^2)...}

F=(9×10^-18){(1+1/9^2+1/9^4+..)-1/3^2(1+1/9^2+1/9^4+..)}

F=(9×10^-18)(1-1/3^2)(1+1/9^2+1/9^4+..)

F=(9×10^-18)(8/9)(1+1/9^2+1/9^4+..)

F=(10^-18)(8)(1+1/9^2+1/9^4+..)

Now using gp formula for infinite series i.e S= (1+1/9^2+1/9^4+..) we get S= (1/(1-(1/9^2)))=9^2/80

F=9^2×(10^-18)(8)/80

F=81×10^-17