Asked by Sumit
Two positive point charges Q and Q’ having charges of +4 μC and +1 μC respectively are fixed at a distance 2 m apart in vacuum.A third point charge q having a charge +0.2 μC is to be placed between Q and Q’, at a distance x from Q’. If q remains at rest, what is the value of distance x?
Answers
Answered by
Sumit
I know I need to use columns law to do this but I don’t know how to proceed.
Answered by
Damon
Coulomb's
!
+4*10^-6 C at x = 2
+1*10^-6 C at x =0 note locations reversed so x is distance from Q'
it does not matter what the value of Q between is, we must find where E = 0 so call it Q
k *1*Q /x^2 = k * 4 * Q/ (2-x)^2
4 x^2 = (2-x)^2
4 x^2 = 4 - 4 x + x^2
3 x^2 + 4 x - 4 = 0
(3x-2)(x+2) = 0
positive x = 2/3 meter = distance from smaller charge
!
+4*10^-6 C at x = 2
+1*10^-6 C at x =0 note locations reversed so x is distance from Q'
it does not matter what the value of Q between is, we must find where E = 0 so call it Q
k *1*Q /x^2 = k * 4 * Q/ (2-x)^2
4 x^2 = (2-x)^2
4 x^2 = 4 - 4 x + x^2
3 x^2 + 4 x - 4 = 0
(3x-2)(x+2) = 0
positive x = 2/3 meter = distance from smaller charge
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