Find q1 such that Force on Test Charge has no X-Component

AI Thread Summary
To ensure the force on a test charge at (5,0,6) has no x-component, the electric field contributions from point charges q1 and q2 must balance in the x-direction. Given q2 is 4nC at (2,0,1), the position of q1 at (4,0,-3) can be adjusted by calculating the required value for q1. The test charge's nature does not affect the calculations, as any non-zero charge will suffice. The focus is on determining the correct magnitude and sign of q1 to achieve the desired force condition. The solution requires applying principles of electrostatics to find the appropriate value for q1.
zekester
Messages
30
Reaction score
0
There are two point charges q1 located at (4,0,-3) and q2 at (2,0,1) if q2=4nC find q1 such that the force on a test charge at (5,0,6) has no x-component. I think I would know how to do this problem if I knew what a test charge was. Is it just one electron.
 
Physics news on Phys.org
The point is that doesn't matter what the charge is (as long as it's not zero). If you were to place a charge q_3 at (5, 0, 6) what would q_1 have to be in order that the force acting on q_3 due to the electric field produced by q_1 and q_2 would give zero x component.
 
thanks for your help
 
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}...
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...
Back
Top