Question

An attractive force of 7.2 N occurs between two point charges that are 0.10 m apart. If one charge is -4.0 µC, what is the other charge? (µC = 1.0 × 10^-6 C)

-4.0 µC , -2.0 µC , +2.0 µC , or +4.0 µC

Answer 1

i think it's C.

Answer 2

here we have to apply teh law of Coulomb

Answer 3

the*

Answer 4

Not so sure! sorry!

Answer 5

\[\Large F = K\frac{{{Q_1}{Q_2}}}{{{d^2}}}\]

Answer 6

+2.0 µC

Answer 7

okay! what do we plug in?

Answer 8

for example I will solve that fromula for Q2, and I get:
\[{Q_2} = \frac{{F{d^2}}}{{K{Q_1}}}\]

Answer 9

7.2 should be plugged in

Answer 10

ohh what would that look like? :/ i am confused :/ is is 7.2 divided by somehing?

Answer 11

substituting our data, we can write:
\[\Large {Q_2} = \frac{{F{d^2}}}{{K{Q_1}}} = \frac{{7.2 \times {{\left( {{{10}^{ - 1}}} \right)}^2}}}{{9 \times {{10}^9} \times \left( { 4 \times {{10}^{ - 6}}} \right)}}\]

Answer 12

F = k*q1*q2/r^2 so q2 = F*r^2/(k*q1) = 7.2*0.10^2/(9.0x10^9*4.0x10^-6) = 2.0x10^-6 and since the force is attractive the charge must be
C - 2.0 µC

Answer 13

sorry b! not c!

Answer 14

no, it is C, since the force is attractive, and only opposite charges interact with an attractive force

Answer 15

ohh so our solution is choice C?

Answer 16

yes!

Answer 17

yay! thank you both!!

Answer 18

your welcome!

Answer 19

was that your last question?

Answer 20

oh never mind