E field problem, just a simple question

  • Thread starter Thread starter pyroknife
  • Start date Start date
  • Tags Tags
    Field
AI Thread Summary
To find the value of charge q' that makes the electric field at the center of the triangle zero, the signs of the charges do matter. The relationship between the charges can be expressed as a multiple of q, simplifying the calculation. By focusing on the magnitudes and the arrangement of the charges, one can derive q' without needing to specify the sign of q. The configuration of the charges at the triangle's vertices and midpoint is crucial for determining the net electric field. Ultimately, understanding the charge interactions is key to solving the problem effectively.
pyroknife
Messages
611
Reaction score
4
Two point particles, each having a charge equal to q, sit on the base of an equilateral triangle that has sides of length L as shown in the figure below. A third point particle that has a charge equal to Q = 5q sits at the apex of the triangle. A fourth point particle that has charge q' is placed at the midpoint of the baseline making the electric field at the center of the triangle equal to zero. What is the value of q' ? (The center is in the plane of the triangle and equidistant from all three vertices.)




My question is, if you want to find the value of q' don't you have to know the signs of some of the charges?
 
Physics news on Phys.org
I think you should determine the answer as a multiple of q. If you do that, you don't need to know the sign of q.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top