Electric Dipole, net force and torque.
I have attached a diagram.
s
Three charge are at corners of an triangle. The + and - 5.00*10^-6C charges form a dipole.
a) Find the magnitude and direction of the force exerted on the dipole from the -10*10^-6C charge.
b) For an axis perepndiculat to the line connected the + and - 5.00*10^-6C charges , find the torque exerted on the dipole by the -10*10^-6 C charge, direction and magnitude?
My attempt:
a) The net force = 0.
Its magnitude on both charges = 1125 N, (from Coloumb's law and using r as 2cm.
However the answer is 1680N on both charges respectively.
b) Torque = Force * distance perpendicular to the line of action.
Considering a single charge, the force exerted on it is 1125N, and the distance perpendicular to the line of action 1.5cm.
The net torque will then = 2(1.5*10^-2 * 1125) , as the force acts in opposite directions on both charges, consequently rotating the dipole system in the same way.
T = 33.75N.m
(The correct answer is however 22.3N.m)
Thanks guys, appreciated
s
Three charge are at corners of an triangle. The + and - 5.00*10^-6C charges form a dipole.
a) Find the magnitude and direction of the force exerted on the dipole from the -10*10^-6C charge.
b) For an axis perepndiculat to the line connected the + and - 5.00*10^-6C charges , find the torque exerted on the dipole by the -10*10^-6 C charge, direction and magnitude?
My attempt:
a) The net force = 0.
Its magnitude on both charges = 1125 N, (from Coloumb's law and using r as 2cm.
However the answer is 1680N on both charges respectively.
b) Torque = Force * distance perpendicular to the line of action.
Considering a single charge, the force exerted on it is 1125N, and the distance perpendicular to the line of action 1.5cm.
The net torque will then = 2(1.5*10^-2 * 1125) , as the force acts in opposite directions on both charges, consequently rotating the dipole system in the same way.
T = 33.75N.m
(The correct answer is however 22.3N.m)
Thanks guys, appreciated
bump?
a) There is a net force on the dipole in the downwards direction. Imagine vectors representing the attraction and repulsion forces corresponding to each end of the dipole and add them. What you need to do is work out the vertical component of the force on each dipole and add them. Then you should get the right answer.
b) Here you have again failed to look at the components correctly. In a) you needed the component of the force in the vertical direction, now you need it in the horizontal direction (ie perpendicular to the axis of interest).
It is probably easiest here to think of torque as the distance from the axis (in this case the line perpendicular to the line joining the charges) multiplied by the component of the force perpendicular to the axis. This is equivalent to the definition you have given.
Really, the easiest way to think about torques is in terms of vector products, but if you haven't encountered them yet don't worry as the simplest mechanics/electromagnetism problems don't require them.
b) Here you have again failed to look at the components correctly. In a) you needed the component of the force in the vertical direction, now you need it in the horizontal direction (ie perpendicular to the axis of interest).
It is probably easiest here to think of torque as the distance from the axis (in this case the line perpendicular to the line joining the charges) multiplied by the component of the force perpendicular to the axis. This is equivalent to the definition you have given.
Really, the easiest way to think about torques is in terms of vector products, but if you haven't encountered them yet don't worry as the simplest mechanics/electromagnetism problems don't require them.
Original post by notastampcollector
a) There is a net force on the dipole in the downwards direction. Imagine vectors representing the attraction and repulsion forces corresponding to each end of the dipole and add them. What you need to do is work out the vertical component of the force on each dipole and add them. Then you should get the right answer.
b) Here you have again failed to look at the components correctly. In a) you needed the component of the force in the vertical direction, now you need it in the horizontal direction (ie perpendicular to the axis of interest).
It is probably easiest here to think of torque as the distance from the axis (in this case the line perpendicular to the line joining the charges) multiplied by the component of the force perpendicular to the axis. This is equivalent to the definition you have given.
Really, the easiest way to think about torques is in terms of vector products, but if you haven't encountered them yet don't worry as the simplest mechanics/electromagnetism problems don't require them.
b) Here you have again failed to look at the components correctly. In a) you needed the component of the force in the vertical direction, now you need it in the horizontal direction (ie perpendicular to the axis of interest).
It is probably easiest here to think of torque as the distance from the axis (in this case the line perpendicular to the line joining the charges) multiplied by the component of the force perpendicular to the axis. This is equivalent to the definition you have given.
Really, the easiest way to think about torques is in terms of vector products, but if you haven't encountered them yet don't worry as the simplest mechanics/electromagnetism problems don't require them.
Ahhh of course. ta. sorted now.
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