A point charge q1 = 4.05 is placed at the origin, and a second point charge = -2

Question

A point charge q1

= 4.05

is placed at the origin, and a second point charge

= -2.90nC

is placed on the x

-axis at x=+

20.5cm

. A third point charge q3

= 2.10nC

is to be placed on the

-axis between q1

and q2

. (Take as zero the potential energy of the three charges when they are infinitely far apart.)

a) What is the potential energy of the system of the three charges if q3

is placed at x=

10.0cm

?

b)Where should q3

be placed between q1

and q2

to make the potential energy of the system equal to zero?

Explanation / Answer

energy=(1/4*pie*epsilon)((q3 q2/r32)+(q3q1/r31)+(q2q1/ r21))

r32=distance between q3 and q2=20.5-10=10.5cm

r31=distance between q3 and q1=10-0=10cm

r21=distance between q2 and q1=20.5-0=20.5cm

substituting values we get

energy=(-5.8*10^-17+8.50*10^-17-5.72*10^-17)*8987551788=-0.27*10^-6 joules





the energy between q1 and q2= -5.72*10^-17*8987551788=-0.00000051408joules----(3)

let x coordinate of q3=xcm

the energy between q1 and q3=(4.05*10^-9*2.1*10^-9*10^2/x)*8987551788---(1)

the energy between q3 and q2=(-2.9*10^-9*2.1*10^-9*10^2/x-20.5)*8987551788----(2)

1+2+3=0

solving we get

x=37.98cm or 8.01cm

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