Two test charges suspended in vacuum attract each other with a force of 1x10^-6 N. What will the attractive force be if the vacuum is replaced with water at 20 C? The permittivity of water at this temperature is 7.07x10^ -10 C^2/N-m^2.(adsbygoogle = window.adsbygoogle || []).push({});

Fwater = (1/4pi*Ewater)(Q1Q2/r^2)

Force is in Newtons

Ewater is the permittivity of water

Q1 and Q2 are the charges in Coulombs

r is the separation in meters

My attempt :(

I am told that this can be simplified down to the following equation.

Fwater = K / Fwater

K is constant, calculated by rolling everything that doesn't change (4pi, the charges, the distance) into one number.

K = 1/4pi*Ewater

K = 1/1.256^-1*7.07 x 10^-10 C^2/N-m^2

K = 1/8.88^-11 C^2/N-m^2

K = 1.12^-10 C^2/N-m^2

After finding K is this the correct formula?

Fwater = (1.12^-10 C^2/N-m^2) / (7.07 x 10^-10 C^2/N-m^2)

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# Homework Help: Attractive Force of two charges in water

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