Electric Field and Potential on a Line with Opposite Charges

  • Thread starter Thread starter silver_gry
  • Start date Start date
  • Tags Tags
    Electric
AI Thread Summary
A positive charge +q is positioned 3.00m to the left of a negative charge -q2, creating a scenario where the net electric field is zero 1.00m to the right of the negative charge. To solve for the two spots where the electric potential is zero, one must equate the electric fields from both charges, leading to a relationship between their magnitudes. By selecting a point on the line through the charges and calculating the potential as the sum of contributions from each charge, the problem can be simplified. Substituting the relationship derived from the electric field equation allows for the cancellation of variables and facilitates finding the locations of zero potential. The solution ultimately reveals the positions of these points relative to the negative charge.
silver_gry
Messages
3
Reaction score
0

Homework Statement


A positive charge of +q is located 3.00m to the left if a negative charge -q2. The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot 1.00m to the right of the negative charge. On this line there are also two spots where the potential is zero. Locate these 2 spots relative to the negative charge.


Homework Equations



I don't know which equation I should use to start with. I really need help with this question.

The Attempt at a Solution

 
Physics news on Phys.org
At the point where E = 0, the E field due to the positive charge is equal (and opposite) to the E field due to the negative charge. Write this, replace each E with the formula for E, put in the numbers for the distance. You will end up with Q1 = a number times Q2.

Now pick a point x on the line through the charges. Say, x meters to the right of Q1. Write the expression for the potential there - the sum of the potentials due to each charge. Use the multiple of Q1 in place of Q2 so the Q1 cancels out in this equation. You should be able to solve it and find x.
 
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