Question

Please help me out. :)

Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsion force of 95 N. What is the magnitude of the larger charge?

Answer 1

use coulomb's law
\[F = k\dfrac{q_1q_2}{r^2}\]

Answer 2

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgele/coulaw.gif

Answer 3

ahh.. okay.. I'll try to solve. :)

Answer 4

just plugin the given values after converting to appropriate units

k = 9*10^9
F = 95
r = 15*10^-2

Answer 5

Q1Q2 = 2.375 x 10^-10 C^2

Answer 6

looks good!

Answer 7

Q1 = Q
Q2 = 2Q

Answer 8

yes solve Q, just algebra..

Answer 9

Ahh.. But I don't know how to separate Q1 and Q2 in the equation.. How will I do that sir? :)

Answer 10

Notice Q1 = Q2/2

Answer 11

Q1Q2 = 2.375 x 10^-10 C^2


becomes


(Q2/2)Q2 = 2.375 x 10^-10 C^2

Answer 12

Notice Q1 = Q2/2 .. How come?

Answer 13

Ahh.. Got it... :) Thanks. :)

Answer 14

We're given one charge is twice as the other
that means the smaller charge is half of the bigger charge, yes ?

Answer 15

Q1Q2 = 2.375 x 10^-10 C^2


becomes


(Q2/2)Q2 = 2.375 x 10^-10 C^2

Q2^2/2 = 2.375 x 10^-10 C^2

Q2^2 = 2*2.375 x 10^-10 C^2

Q2 = sqrt(2*2.375 x 10^-10 )

Answer 16

http://www.wolframalpha.com/input/?i=sqrt%282*2.375+x+10%5E%28-10%29+%29

Answer 17

Thank you! :) I understand it well. :)