A charge Q exerts a 24 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q?
A) 3 N
B) 6 N
C) 24 N
D) 36 N
E) 48 N
Best answer gets the points.
2006-12-22 16:01:09 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
6 N....
F= q x q/ r^2
according to the data.... F = 24 N
that implies that 24= q x q / r^2
if distance is doubled = 2r,
then square of the distance = 4 r^2
that implies .... q x q / 4r^2 =24
but q x q/ r^2 = 24 so substituting in second equation we get....
24/4 = F
therefore force F = 6 N
2006-12-22 16:22:12 · answer #1 · answered by Hawk 2 · 0⤊ 0⤋
B) 6N
The equation is F = k(Q1xQ2)/(R^2), where k is a constant, Q1 and Q2 are the charges and R is the distance between them.
If R is doubled, the overall equation is divided by 4. So the original force of 24 N is divided by 4, and becomes 6N.
2006-12-22 16:09:48 · answer #2 · answered by Sephisabin 3 · 0⤊ 0⤋
the equation F=kq1q2/r^2. the numerator can be ignored since the constant(k) and the two charges remain constant, and the distance doubles so the charge is quartered which is choice B, 6 Newtons
2006-12-22 16:08:28 · answer #3 · answered by jdog33 4 · 0⤊ 0⤋
F= k*q1*q2/r^2
Put r= 2r in the equation
Thus F2=1/(4*F1)
F2=24/4=6
Force is quartered.
B)6 N
2006-12-22 16:07:15 · answer #4 · answered by Som™ 6 · 0⤊ 0⤋
all my above answeres have said a lot already...
i can only reassure u that it is 6N all right...
2006-12-22 18:24:21 · answer #5 · answered by catty 4 · 0⤊ 0⤋
no..it was 3N
2016-01-24 17:04:17 · answer #6 · answered by nur syakira 1 · 0⤊ 0⤋