Electric field of point along the central vertical axis of a triangle

[Samir_Khalilullah]

Homework Statement

Please see attachments.

Relevant Equations

$E = \frac {k q} {r^2}$
$F= Eq$
$\ddot x + \omega^2 x = 0$

here is my attempted solution.

$d^2 = z^2 + \frac {L^2} {3}$
$C$ is coulomb constant
since the point is symmetric, only the vertical component of the electric field remains. So,
$$ E = 3 E_y =3 \frac {C Q cos \theta} {d^2} $$
$$ E= 3 \frac {C Q z} {d^3} $$

thus part (a) is done ( i think).

for part (b),

the particle with mass $m$ and charge $-q$ is experiencing a force

$$ F = -Eq = \frac {-3 C Q q z} {d^3} $$
$$ m \ddot z + \frac {3 C Q q z} {d^3} = 0 $$
$$ \ddot z + \frac {3 C Q q z} {m d^3} = 0 $$
Comparing it with the SHM equation ,

$$ \omega^2 = \frac {k_{spring}} {m} = \frac {3 C Q q } {m d^3} $$

now we know,
$$ K_{energy} = \frac {1} {2} k_{spring} x^2$$
thus we get ,
$$ x = \sqrt{\frac {2K_{energy} d^3} {3 C Q q} } $$

Did i do it correctly??
If i did then im going to approach part (c)

 

[kuruman]

People will be more likely to make an effort to help you if you made an effort to put the statement of the problem where it belongs, in the template instead of in attachments.

 

[Samir_Khalilullah]

People will be more likely to make an effort to help you if you made an effort to put the statement of the problem where it belongs, in the template instead of in attachments.

sorry ... Should i post a new thread on this with the statement on the template?

 

[Steve4Physics]

Homework Statement: Please see attachments.

For information, note that the question states that the reduced Planck constant ($\hbar$) is $6.62 \times 10^{-34}$ Js. That is wrong because $\hbar = \frac h{2\pi}$ and $h=6.63 \times 10^{-34}$ Js (not even $6.62 \times 10^{-34}$ Js). However the Planck constant is not required in the question!!!

It may also be worth noting that system described is unstable (c.f. a sharp pencil balanced on its tip). The negative charge will be attracted to one of the positive charges unless it is constrained to the symmetry-axis. (As explained by Earnshaw’s theorem if you want to look it up.)

$C$ is coulomb constant

I'd stick to $k$. It is distinguishable from $K$ and (if needed) from $k_{spring}$.

$$ E= 3 \frac {C Q z} {d^3} $$
thus part (a) is done ( i think).

Both $z$ and $d$ are variables. You've already said $d^2 = z^2 + \frac {L^2} {3}$. Your equation for $E$ should be in terms of the single variable $z$.

for part (b),

the particle with mass $m$ and charge $-q$ is experiencing a force

$$ F = -Eq = \frac {-3 C Q q z} {d^3} $$
$$ m \ddot z + \frac {3 C Q q z} {d^3} = 0 $$
$$ \ddot z + \frac {3 C Q q z} {m d^3} = 0 $$
Comparing it with the SHM equation ,

$$ \omega^2 = \frac {k_{spring}} {m} = \frac {3 C Q q } {m d^3} $$

The equation for SHM needs to be in terms of a single variable, $z$. An equation for $\omega$ should be in terms of constants, but $d$ is not a constant'

You need a differential equation with a single variable ($z$), not both $z$ and $d$. The equation may be a little messy but when you consider small values of $z$, you should be able to make an approximation to simplify the equation to the 'standard' SHM equation.

An alternative approach might be to consider energy conservation.

$$ x = \sqrt{\frac {2K_{energy} d^3} {3 C Q q} } $$

You are asked for a value of $z$, not '$x$'. And the question refers to this value as 'D'. So your final equation should start 'D =', not 'x ='.