Find the Potential as a function of position

AI Thread Summary
To find the electric potential as a function of position along the x-axis due to a uniformly charged rod along the y-axis, one must integrate the potential contributions from each infinitesimal charge element. The potential is expressed as V(x) = ∫(0 to L) (k/r) dq, where r is the distance from the charge element to the point on the x-axis. The charge element dq can be expressed in terms of dy, and r can be determined using the Pythagorean theorem. Proper integration will yield the desired potential function. Understanding these relationships is crucial for solving the problem effectively.
prace
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Homework Statement


A rod of length L carries a charge Q uniformly distributed along its length. The rod lies along the y-axis with one end at the origin. Find the potential as a function of position along the x-axis


Homework Equations


dV=\vec{E}\cdotp d\vec{l}

V=\frac{kq}{r}


The Attempt at a Solution



I think I am to use the first equation posted, but I am not sure how to relate it to the x-axis, if the rod lies along the y-axis. Also, what is the displacement here? I am assuming it is just dy?
 
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prace said:

Homework Statement


A rod of length L carries a charge Q uniformly distributed along its length. The rod lies along the y-axis with one end at the origin. Find the potential as a function of position along the x-axis

Homework Equations


dV=\vec{E}\cdotp d\vec{l}

V=\frac{kq}{r}

The Attempt at a Solution



I think I am to use the first equation posted, but I am not sure how to relate it to the x-axis, if the rod lies along the y-axis. Also, what is the displacement here? I am assuming it is just dy?
You have to integrate along the rod from y = 0 to +L.

V(x) = \int_{0}^{L} \frac{k}{r}dq

Work out the expression for dq in terms of dy. Work out the expression for r in terms of x and y (think: Pythagoras).

AM
 
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Thread 'Variable mass system : water sprayed into a moving container'

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}...
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