To find the distance from charge q1 where the electric field is zero between two positive charges q1 and q2 separated by distance s, the correct approach involves setting the magnitudes of the electric fields from both charges equal. The derived equation is x = s / (√(q2/q1) + 1), where x represents the distance from q1. The initial attempt at the solution was incorrect as it solved for s instead of x. The final answer aligns with the calculations provided, confirming its correctness. This method effectively determines the point of zero electric field between the two charges.
#1
electricman
14
0
Homework Statement
Two particles with positive charges q1 and q2 are separated by a distance s.
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero?
Homework Equations
Express your answer in terms of some or all of the variables s , q1 , q2 and k=1/(4pi epsilon)
The Attempt at a Solution
I got s=sqrt(q1(d-s)^2 / q2), but i think its wrong.
Simplify x/x to 1 and add to other side then multiply by x:
<br />
s = (\sqrt{\frac{{q}_{2}}{q}_{1}} + 1)x<br />
Finally divide to get x alone:
<br />
x = \frac{s}{\sqrt{\frac{{q}_{2}}{q}_{1}} + 1}<br />
So that's my final answer, which I'm pretty sure is correct but then I usually make some careless error. (Sorry if I showed too many baby steps with the algebra there.)
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Figure 1 Overall Structure Diagram
Figure 2: Top view of the piston when it is cylindrical
A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.
Figure 3: Modifying the structure to incorporate a fixed internal piston
When I modify the piston...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...