Comparing Electric Field Strengths at Two Points: Positive and Neutral Charges

AI Thread Summary
The discussion centers on comparing electric field strengths at two points influenced by multiple positive charges. Point 1, with four positive charges, initially seems stronger, but the repulsive forces between the charges reduce its effective field strength. In contrast, Point 2 benefits from the additive effect of the charges, resulting in a greater electric field strength. The calculations show that Point 2 has a total field strength of approximately 3.111q, while Point 1 effectively has a reduced strength of 2q. Thus, Point 2 has a stronger electric field than Point 1 due to the interaction of the charges.
dekoi
I am posed with a question, there are two charges, both positive, and two neutral points.

I am asked to compare the electric field strengths at the two points.

Here is a pictorial representation:

\bigoplus------->\bullet 1<-------\bigoplus\bigoplus\bigoplus------->\bullet 2

The three +s in a row represent one big positive charge.

So I drew my electric field lines (represented by the arrows), and now I have to analyze whether the field strengths at the two points are different, or equal.

At first I thought that 1 > 2, since 1 has respectively four positive charges at it, while 2 only has three positive charges.
However, then I thought:
In the case of 1, the two electric fields will repel each other, and then there will no ( ? ) field lines around 1, in which case 2 > 1.

But I don't have much proof to back up my answer, and my textbook doesn't explain this in great detail (neither did my teacher). Any suggestions on how to compare electric field strengths?

Thanks.
 
Physics news on Phys.org
If the two points are of equal distance to the big charge then the field at point 2 will be bigger, here's why:
the total field at point 1 is (k/r^2)*3q - (k/r^2)*q = (k/r^2)*2q the (-) is to show that they are in oppisite directions
the total field at point 2 is (k/r^2)*3q + (k/9r^2)*q = (k/r^2)*3.111q
anyway in point 1 the small charge takes away from the big one, at point 2 it adds to it.
 
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