What is the force, in terms of Coulomb's constant, between two electrical charges of $6 C$ $6 C$ and $- 6 C$ $-6 C$ that are $12 m$ $12 m$ apart?

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Answer 1 · 2017-06-03T02:54:01

$F = - \frac{1}{4} k_{e}$ $F = -1/4k_e$

The electric force $F$ $F$ between two point charges $q_{1}$ $q_1$ and $q_{2}$ $q_2$ a distance $r$ $r$ apart is given by the equation

$F = k_{e} \frac{q_{1} q_{2}}{r^{2}}$ $F =k_e(q_1q_2)/(r^2)$

where $k_{e}$ $k_e$ is Coulomb's constant, equal to $8.988 \times 10^{9} \frac{{N ∙ m}^{2}}{C^{2}}$ $8.988 xx 10^9 ("N • m"^2)/("C"^2)$

Plugging in known values, we have

$F = k_{e} \frac{(6 C) (- 6 C)}{{(12 m)}^{2}}$ $F =k_e((6"C")(-6"C"))/((12"m")^2)$

$= k_{e} \frac{- 36 C^{2}}{144 m^{2}}$ $= k_e(-36"C"^2)/(144"m"^2)$

$= - \frac{1 C^{2}}{4 m^{2}} k_{e}$ $= -(1"C"^2)/(4"m"^2)k_e$

So, in terms of Coulomb's constant ( $k_{e}$ $k_e$ ), the (attractive) electric force acting between the two point charges is

$F = - \frac{1}{4} k_{e}$ $color(red)(F = -1/4k_e)$

(excluding units)

What is the force, in terms of Coulomb's constant, between two electrical charges of 6 C6C6 C and -6 C−6C-6 C that are 12 m 12m12 m apart?