Two charges of $-7 C$ and $4 C$ are at points $(4, 7 ,-8)$ and $( 2 ,-3, -8 )$ , respectively. Assuming that both coordinates are in meters, what is the force between the two points?

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Answer 1 · 2017-07-10T22:30:46

$F_e = 2.42 xx 10^9$ $"N"$

We're asked to find he magnitude of the electric force $F_e$ between two point charges, which we'll call $q_1$ and $q_2$ .

To do this, we can use the equation

$F_e = k(|q_1q_2|)/(r^2)$

where

This distance can be found via the distance formula:

$r = sqrt((4-2)^2 + (7-(-3))^2 + (-8-(-8))^2) = color(red)(10.2$ $color(red)("m"$

Plugging in known values, we have

$F_e = (8.988 xx 10^9("N"•cancel("m"^2))/(cancel("C"^2)))((|(-7cancel("C"))(4cancel("C"))|)/((color(red)(10.2)cancel(color(red)("m")))^2))$

$= color(blue)(2.42xx10^9$ $color(blue)("N"$

Two charges of -7 C -7 C and 4 C 4 C are at points (4, 7 ,-8) (4, 7 ,-8) and ( 2 ,-3, -8 ) ( 2 ,-3, -8 ), respectively. Assuming that both coordinates are in meters, what is the force between the two points?