Two negative charges of unit magnitude and a positive charge q are placed in a straight line. At what position and for what value of q will be the system be in equilibrium
2007-01-04 01:57:17 · 6 answers · asked by GUNA S 1 in Science & Mathematics ➔ Physics
Let distance between q and unit negative charges be x units
Considering symmetry the positive charge q is exactly in between the two unit -ve chargs.
Force of attraction between q and unit -ve charge =
Force of repulsion between the negative charges
k*q/x^2 = k*1/(2x)^2
q = +1/4 units of charges
If distance is in meters and proportionality constant is in SI units then q = 0.25 C
2007-01-04 02:03:27 · answer #1 · answered by Som™ 6 · 0⤊ 0⤋
Som is essentially right.
The positive charge is 1/4 of unit magnitude of one of the negative charges. This positive charge is placed in the middle of the distance separating the 2 negative charges.
2007-01-04 02:43:37 · answer #2 · answered by answerING 6 · 0⤊ 0⤋
+1/6
2007-01-04 02:03:51 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1 st solution is correct But ! it is only one of the solutions
you can find infinitely many different Q and X values
clue:
if a= b c then
a= b/2 c*2
also
a=b/3 c*3
.
.
.
2007-01-04 02:46:48 · answer #4 · answered by come2turkey:) 2 · 0⤊ 0⤋
1) 1st case:- =>By F = kq1q2/r^2 =>at mid point for a unit charge, by F = qE=>F = E as q = 1 =>E = kq^2/r^2 =>1780 = 9 x 10^9 x q^2/(16.7 x 10^-2)^2 =>q = √[5.52 x 10^-9] =>q = 0.74 x 10^-6 C =>q = 0.74 µC 2) It will be zero, as the net force at the mid point on a unit charge will be zero=>By F = qE, if F=0=>E=0
2016-05-23 02:28:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋
in greenwick city
2007-01-04 02:06:25 · answer #6 · answered by Haris K 2 · 0⤊ 0⤋