Question #12904

1 Answer
Oct 10, 2017

#d = 2# m

Explanation:

First, we'll start with the equations:

#F = k (q*Q)/d^2#

and

#E = F/d#

Plugging #F# into our #E# equation, we get:

#E = F/q = {(k*cancelq*Q)/d^2]/cancelq = (k*Q)/d^2#

Where #k# is Coulomb's law constant: #8.99*10^9 (N*m^2)/C^2#
...or more simply: #9*10^9 (N*m^2)/C^2#

Convert to the appropriate units:

#muC => C# and #(kN)/C => N/C#

#E = (k*Q)/d^2 => (2.5X10^3 N/C) = [(9*10^9 (N*m^2)/C^2)(1*10^-6C)]/d^2#

Solve for #d#:

#d = sqrt([(9*10^9 (N*m^2)/C^2)(1*10^-6C)]/(2.5X10^3 N/C)) = 1.89 m#

#=underline2# meters for significant figures (#underline1muC#)