Consider a system of three point charges ont the x axis. Charge 1 is at x=0, charge 2 is at x=0.20m, and charge 3 is at x = 0.40m. In addition, the charges have the following values: q1 = -19uC, q2=q3=+19uC. Find the the point where E =0 between x =0.20m?

Find the point where E =0 between x = 0.20m and x = 0.40m

1 Answer
P dilip_k
Mar 22, 2017

drawn

The situation as described in the problem is shown in the figure.
Let the position where the system charges will have net net intensity #E=0# is the point #N# which is #x# m from #q_2#.

Intensity at N due to #q_1#along #vec(NO)=E_1=k_cxx(19xx10^-6)/(0.2+x)^2NC^-1#.

Intensity at N due to #q_3#along #vec(NO)=E_3=k_cxx(19xx10^-6)/(0.2-x)^2NC^-1#.

Intensity at N due to #q_2#along #vec(NQ)=E_2=k_cxx(19xx10^-6)/x^2NC^-1#.

where Coulomb's constant, #k_c=9xx10^9Nm^2"/"C^2#.

So considering the equilibrium we can write

#E_1+E_3=E_2#

#=>k_cxx(19xx10^-6)/(0.2+x)^2+k_cxx(19xx10^-6)/(0.2-x)^2=k_cxx(19xx10^-6)/x^2#

So we have

#1/(0.2+x)^2+1/(0.2-x)^2=1/x^2#

#=>(2xx0.2^2+2xx x^2)/(0.04-x^2)^2=1/x^2#

#=>(0.08+2x^2)/(0.04-x^2)^2=1/x^2#

#=>(0.08+2x^2)x^2=(0.04-x^2)^2#

#=>2x^4+0.08x^2=x^4-0.08x^2+16xx10^-4#

#=>x^4+2xx0.08x^2+(0.08)^2=16xx10^-4+(0.08)^2#

#=>(x^2+0.08)^2=80xx10^-4#

#=>x^2=sqrt(80xx10^-4)-0.08~~0.00944#

#=>x~~sqrt0.00944~~0.09 m#

So location of #N# on x-axis where #E=0# is #color(red)((0.29,0)#

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