G(x) is f(x) shifted left by (4 -(-2)) = 6 units.
The value of k is 6.
Answer:
it equals x=10 so i guess it's B
Step-by-step explanation:
<span>Assuming that the particle is the 3rd
particle, we know that it’s location must be beyond q2; it cannot be between q1
and q2 since both fields point the similar way in the between region (due to
attraction). Choosing an arbitrary value of 1 for L, we get </span>
<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>
Rearranging to calculate for d:
<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207 </span>
<span>
We pick the value that is > q2 hence,</span>
d = 2.72075922005613*L
<span>d = 2.72*L</span>
