Does Voltage at Point P Reflect Net Attraction Despite Equal Charges?

  • Thread starter Thread starter mm2424
  • Start date Start date
  • Tags Tags
    Direction Voltage
AI Thread Summary
The discussion centers on calculating the net electrical potential at point P due to four particles, with a specific focus on understanding the nature of voltage as a scalar quantity. It confirms that electrical potential from multiple charges adds together according to the superposition principle, meaning that even if only positive charges are present, their contributions do not cancel out but sum to a positive value. The confusion arises from interpreting the implications of this voltage in terms of net attraction, as the electrical fields from equal positive charges do not negate each other. Ultimately, the calculated potential of 0.54 mV indicates a net positive voltage at point P, affirming that the contributions from the positive charges accumulate rather than cancel. This reinforces the understanding that potential is additive and does not imply a net attraction from equal charges.
mm2424
Messages
43
Reaction score
0

Homework Statement



What is the net electrical potential at point P due to the four particles if V = 0 at infinity, q = 5 fC and d = 4 cm? The image for this problem can be seen at http://www.webassign.net/hrw/hrw7_24-33.gif

Homework Equations



Potential = 1/(4εpi) * q/r

The Attempt at a Solution



I was able to answer this question, and the answer agrees with the answer key. However, it's left me a bit confused. Potential is a scalar, correct? How does one interpret the meaning of the voltage for point P if, say, only the two +q charges were present. Do the voltages cancel out, or do they add up? I ask because, when I answered this problem, I just blindly added up all the potentials at point P from each charge, regardless of the charges' location. I added the pluses, subtracted the minuses, and got the right answer (which ends up being .54 mV). But does that mean that there is a net positive attraction for the particle? I don't see how that makes sense, since the electrical fields from the positive charges should cancel, no? I imagine I'm looking at this the wrong way...any help would be greatly appreciated!
 
Physics news on Phys.org
mm2424 said:
I was able to answer this question, and the answer agrees with the answer key. However, it's left me a bit confused. Potential is a scalar, correct? How does one interpret the meaning of the voltage for point P if, say, only the two +q charges were present. Do the voltages cancel out, or do they add up?

The potentials add. Potential obeys the superposition rule, so you can treat one sourtce at a time and sum the contributions of each.
 
So if you had a point that was equidistant between two equal positive charges q, the voltage would be positive and not 0?
 
mm2424 said:
So if you had a point that was equidistant between two equal positive charges q, the voltage would be positive and not 0?

Yes. The potential (a scalar) would be the sum of the contributions from each charge, both of which are positive.
 
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