Question

An attractive force of 7.2 N occurs between two point charges that are 0.05 m apart. If one charge is - 4.0 µC, what is the other charge?

A. +0.5 µC
B. +2.0 µC
C. -2.0 µC
D. -4.0 µC

Answer 1

@Michele_Laino :)

Answer 2

here we have to apply the law of Coulomb:
\[\Large F = K\frac{{{q_1}{q_2}}}{{{d^2}}}\]

Answer 3

ok! what do we plug in?

Answer 4

do you know for what the symbols stand @iheartfood

Answer 5

yes, i believe so....

i just get confused with inserting :/

Answer 6

oh
tell us what the symbols mean

Answer 7

the next step is:
\[\large {q_2} = \frac{{F{d^2}}}{{K{q_1}}} = \frac{{7.2 \times {{\left( {5 \times {{10}^{ - 2}}} \right)}^2}}}{{9 \times {{10}^9} \times 4 \times {{10}^{ - 6}}}} = ...Coulombs\]

Answer 8

we get 5E-7?

Answer 9

is that right? if so, what is the next step? :)

Answer 10

that's right!

Answer 11

what is the sign of q_2?

Answer 12

positive?

Answer 13

yes! since the force is attractive and q_1 is negative

Answer 14

attractive force exists between opposite charrges

Answer 15

ooh yay! so our solution will be either -2.0 or -4.0 ? how can we find which it is?

Answer 16

our solution is:
\[\large 5 \times {10^{ - 7}} = 0.5 \times {10^{ - 6}} = 0.5\mu C\]

Answer 17

no
if one is negative other will be positive

Answer 18

oh ok! yay thank you!! +0.5 it is!!

Answer 19

yes! @rvc

Answer 20

thanks

Answer 21

that's right! @iheartfood