Kinetic and potential energy of a particle attracted by charged sphere

[fatpotato]

Homework Statement

Find the speed of a particle initially at rest when put next to a charged sphere.

Relevant Equations

Potential energy $U_E = k\frac{q_1 q_2}{r}$
Kinetic energy $E_k = \frac{1}{2}mv^2$

Hello,

I have a particle at point A with charge $q_A$, and an unmovable sphere of radius $R_B$ at point B with a volumic charge density $\rho$. The distance from particle A to the centre of the sphere in B is $r$. Both objects have opposed charges, so, the particle in A, initially at rest, is attracted to the charged sphere and reaches it with a certain speed with modulus $v$, which I have to find.

First, I suppose the charge of the sphere is simply $Q_B = \rho V_B = \rho \frac{4}{3}\pi R_B^3$.

Then, I suppose that the electric potential energy will be entirely converted into kinetic energy $E_k$, when the particle reaches the sphere. I assume (and it might be wrong) that the sphere can be considered as a particle, so I consider that the two objects will touch after the particle travels a distance $r$ although in reality it touches the sphere after traveling a distance $r - R_B$.

In the end, it boils down to $$U_E = E_k \rightarrow k\frac{q_A Q_B}{r} = \frac{1}{2}mv^2$$ So I solve for : $$v = \sqrt{\frac{2kq_AQ_B}{rm}}$$ while removing my negative sign due to the product $q_AQ_B$ to avoid a complex result.

However, my answer differs from the correction. Speed should be $v = 2.5\cdot 10^5 \frac{m}{s}$, but I get $8.42 \cdot 10^4 \frac{m}{s}$ using the following numerical values :
$R=0.1m$, $r = 1m$, $k = \frac{1}{4\pi \varepsilon_0} \approx 9\cdot 10^9$, $\rho = -1 \frac{\mu C}{m^3}$, $m = 1.7\cdot 10^{-27} kg$, $q_A = +1.6 \cdot 10^{-19}C$

Can anyone pinpoint what I am doing wrong please?

Thank you

 

[PeroK]

In the end, it boils down to $$U_E = E_k \rightarrow k\frac{q_A Q_B}{r} = \frac{1}{2}mv^2$$ So I solve for : $$v = \sqrt{\frac{2kq_AQ_B}{rm}}$$ while removing my negative sign due to the product $q_AQ_B$ to avoid a complex result.

This is not right. When the particle collides with the sphere, it is still a distance $R$ from the centre of the sphere, hence still has some potential energy.

And, in fact, the potential energy is zero at infinity, which is why you had an incorrect negative sign to get rid of!

 

[haruspex]

Further to @PeroK's comment, the potential energy lost is not $\frac{kqQ}r$. That is the PE required to be added to move the particle away to infinity. If we define the PE at infinity as zero (which is usual) then the PE at distance r is $-\frac{kqQ}r$.
If a particle moves from distance r away to distance s away the lost PE is $-\frac{kqQ}r-(-\frac{kqQ}s)=kqQ(\frac 1s-\frac 1r)$.

 

[fatpotato]

Of course!

By taking into account that the particle still has potential energy of $kq_AQ_B \cdot (\frac{1}{r} - \frac{1}{R})$ when touching the sphere I get to the correct result.

Thank you PeroK!

Edit : just saw the answer from haruspex : thank you!

 

[PeroK]

Further to @PeroK's comment, the potential energy lost is not $\frac{kqQ}r$. That is the PE required to be added to move the particle away to infinity. If we define the PE at infinity as zero (which is usual) then the PE at distance r is $-\frac{kqQ}r$.
If a particle moves from distance r away to distance s away the lost PE is $-\frac{kqQ}r-(-\frac{kqQ}s)=kqQ(\frac 1s-\frac 1r)$.

As opposite charges attract, the negative sign is not needed for electrostatic potential.

 

[Delta2]

Further to @PeroK's comment, the potential energy lost is not $\frac{kqQ}r$. That is the PE required to be added to move the particle away to infinity. If we define the PE at infinity as zero (which is usual) then the PE at distance r is $-\frac{kqQ}r$.
If a particle moves from distance r away to distance s away the lost PE is $-\frac{kqQ}r-(-\frac{kqQ}s)=kqQ(\frac 1s-\frac 1r)$.

The potential energy is negative because charges have opposite signs. You don't have to put the minus sign there. If the charges had same sign, the test charge would move by the electric force to infinity, no need to add energy to move it to infinity.

 

[haruspex]

As opposite charges attract, the negative sign is not needed for electrostatic potential.

Yes, I was thinking in terms of gravitational potential energy and failed to consider the charge signs. But the principle that the PE change takes the form 1/s-1/r stands.