# What is the electric force between a +2 µC point charge and a –2 µC point charge if they are separated by a distance of 5.0 cm? Show your work. (µC = 1.0 × 10–6 C)

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### 1 Answer

To solve for the electric force, apply Coulomb's Law formula which is:

`F= | (kq_1q_2)/r^2|`

where F - electric force (in Newton) ,

q1 and q2 - are the charges (in Coulomb),

r - distance between q1 and q2 ( in meters), and

k - electrostatic constant `9xx10^9 N*m^2/C^2` .

So, convert 5.0 cm to meters.

`5.0cmxx(1m)/(100cm)=0.05m`

Also, convert the given charges to Coulombs.

`+2muCxx(1xx10^(-6)C)/(1muC)=+2xx10^(-6)C`

`-2muCxx(1xx10^(-6)C)/(1muC)=-2xx10^(-6)C`

Then, plug-in `k=9xx10^9` , `q_1=+2xx10^(-6)` , `q_1=-2xx10^(-6)` and `r=0.05` to the formula.

`F=|((9xx10^9)(+2xx10^(-6))(-2xx10^(-6)))/0.05^2|`

`F=|-14.4|`

`F=14.4`

**Hence, the electric force between the given charges is 14.4N.**