# Speed at Which Point Charge in Approaching origin / loss of charge at constant rate

a)Point charge Q = +7.25 μC is fixed at the origin. Point charge q = +1.60 μC is now carefully placed on the positive y-axis, and it floats at (0,7.88). Find the mass of q.

b) Refer to question a. Suppose that, as charge q is floating, the point charge Q at the origin begins to lose its charge at a constant rate of dQ/dt = -0.285 μC/s. Find the speed with which point charge q is approaching the origin when it is at (0, 3.94).

I got the answer a by using F = k x (q1 x q2 / r^2) and F = mg ---> m = 1.71 kg

Simon Bridge
Homework Helper

m = 1.71 kg seems a bit on the heavy side to me
coulomb constant is order 1010, the product of the charges is order 10-12 they gets divided by a number to the order of 100 .... expect a mass order of 0.0001kg.

- did you remember the 10-6 on each charge?

For (b) they are saying that conservation of charge is being violated ... it's magically being spirited away[*]. In context, they probably only only want the force to change with time. So you want to write out ƩF(r,t)=mar(t) and solve for r(t) and vr(t) I guess.

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[*] Perhaps the charge at the origin is a charged metal sphere and charge is being drained to ground through a resistor?

1.71 kg is correct.

Still lost on b; F = 16.794 N and it is only in y direction

Simon Bridge
Homework Helper

Still lost on b; F = 16.794 N and it is only in y direction
The coulomb force cannot be constant - Q depends on time
... "only in the y direction" just means that r=yj and, so, r=y ;)
... and don't forget gravity in the -y direction.
1.71 kg is correct.
Since you insist. That y=7.88 distance is not meters then? (Must be cm).

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My profound apologies = Yes, in cm.

Simon Bridge
Homework Helper

I have to avoid doing the problem for you ... so I'll stick to stuff you already know:

Choosing +y as the positive direction for forces:

The coulomb force is (remembering that y is a function of time now) $$F_C=\frac{kqQ(t)}{y^2}:Q(t)=a-bt$$
The gravity force is $$F_g=-mg$$

The net force in the +y direction is mass times y-acceleration$$F_C - F_g = m\ddot{y}$$

:D

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Simon Bridge
Homework Helper

... I don't see how that is "introductory" ... what level is this to be done at?

I suppose you could assert that the coulomb and gravity forces are equal and opposite all through the motion - in which case the speed is a constant.

So the speed is just distance over time ... have the distance, need the time to travel Δy
This is the time it takes for the coulomb force to reduce so it will balance q at half the distance.

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Thank you Simon -- I'll give this a go...

".. I don't see how that is "introductory" ... what level is this to be done at?" = My thoughts exactly...

...nd day of Principles of Physics II...