Understanding Point Charges: Same Sign, Different Magnitudes?

AI Thread Summary
The discussion centers on a problem involving three point charges at the corners of a square, where the goal is to achieve a net electric field of zero at the empty corner. It is clarified that the charges cannot all have the same sign; instead, they must have different signs to balance the electric fields. Specifically, one charge should be of one type (positive or negative) while the other two should be of the opposite type. The direction of the electric field, influenced by the sign of the charges, is crucial for achieving equilibrium. The conclusion emphasizes that understanding the vector nature of electric fields is key to solving the problem correctly.
smunger81
Messages
12
Reaction score
0
Please Help!

I would really appreciate some help on this one...I'm pretty sure I'm making it harder than it is...

The question is this - "Three point charges are fixed to the corners of a square, one to a corner, in such a way that the net electric field at the empty corner is zero. Do these charges all have the same sign or the same magnitude but perhaps different signs?"

My answer is the charges would have the same sign and different magnitudes because when calculating net electric fields the sign of the charge is not taken into account...is this right?

Any advice would be fabulous! Thank you!
 
Last edited:
Physics news on Phys.org
Sorry,
Let the empty corner is the origin and the three other points are (0, -a) (-a, 0) and (-a,-a). If all charges are having same sign (say +) then the forces are along y, x and bisector of x and y respectively. hence the charges must have different sign.

Actually the diagonally opposite corner must have one type of charge and the others have opposite type of charges.

Negative sign of the charge changes the direction of electric field which is a vector quantity.
 
Thank you for your help. I was making it waay to hard.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top