Electrostatic Force. The magnitude of the electrostatic force between two point charges Q and q of the same sign is given by F(x)=kQq / x^2 where x is the distance (measured in meters) between the charges and k=9⋅10^9 Nm^2 / C^2 is a physical constant (C stands for coulomb, the unit of charge; N stands for newton, the unit of force). For two identical charges with Q=q= 2 C, what is the instantaneous rate of change of the force at a separation of x= 0.005 m?

Please show work (if you can).

step-by-step explanation:

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

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