Electrostatic Oscillator

## **Problem**

Two positive point charges

Q

{\displaystyle +Q }

 are fixed to the x-axis at

(
b
,
0
)

{\displaystyle (+b,0) }

, and

(
−
b
,
0
)

{\displaystyle (-b,0) }

. A third charge

q

{\displaystyle +q }

 of mass

m

{\displaystyle m}

, which is free to move along the x-axis, is in equilibrium at

(
0
,
0
)

{\displaystyle (0,0) }

. Ignore the effects of gravity.

**Part 1:** Find an expression for the electric field of this system of charges at point

P
(
x
,
y
)

{\displaystyle P(x, y)}

.

**Part 2:** Show that if

q

{\displaystyle +q }

 is moved by a small amount

x
<<
b

{\displaystyle x<<b }

 away from

(
0
,
0
)

{\displaystyle (0,0) }

 along the x-axis, then released, it will oscillate.

**Part 3:** Find the angular frequency and period of its oscillation.

## **Hint**

(1) Use the binomial expansion

(

1
±

x
b

)

−
2

≈
1
∓
2

x
b

{\displaystyle {\left(1 \pm \frac{x}{b}\right)}^{-2} \approx 1 \mp 2\frac{x}{b} }

 for

x
<<
b

{\displaystyle x<<b }

.

## **Solution**

**Part 1**

There are three point charges along the x-axis; thus, there are three electric field vectors from each point charge to point

P
(
x
,
y
)

{\displaystyle P(x, y)}

:

E
P

→

=

E
1

→

E
2

→

E
3

→

{\displaystyle \vec{E\_P} = \vec{E\_1} + \vec{E\_2} + \vec{E\_3}}

where

E
1

→

=

k
Q

r
1
2

r
1

→

r
1

=

k
Q

[

(
x
b

)
2

y
2

]

3

/

2

[
(
x
b
)
i
y
j
]

E
2

→

=

k
q

r
2
2

r
2

→

r
2

=

k
q

[

x
2

y
2

]

3

/

2

[
x
i
y
j
]

E
3

→

=

k
Q

r
3
2

r
3

→

r
3

=

k
Q

[

(
x
−
b

)
2

y
2

]

3

/

2

[
(
x
−
b
)
i
y
j
]
.

{\displaystyle
\begin{align}
\vec{E\_1} &=\frac{kQ}{r^2\_1} \frac{\vec{r\_1}}{|r\_1|} = \frac{kQ}{\left[(x+b)^2 + y^2 \right]^{3/2}}[(x+b)\boldsymbol{i} + y\boldsymbol{j}] \\[5pt]
\vec{E\_2} &=\frac{kq}{r^2\_2} \frac{\vec{r\_2}}{|r\_2|} = \frac{kq}{\left[x^2 + y^2 \right]^{3/2}}[x\boldsymbol{i} + y\boldsymbol{j}] \\[5 pt]
\vec{E\_3} &=\frac{kQ}{r^2\_3} \frac{\vec{r\_3}}{|r\_3|} = \frac{kQ}{\left[(x-b)^2 + y^2 \right]^{3/2}}[(x-b)\boldsymbol{i} + y\boldsymbol{j}].
\end{align} }

Consequently,

E
P

→

=

[

k
Q
(
x
b
)

[

(
x
b

)
2

y
2

]

3

/

2

k
q
x

[

x
2

y
2

]

3

/

2

k
Q
(
x
−
b
)

[

(
x
−
b

)
2

y
2

]

3

/

2

]

i

[

k
Q
y

[

(
x
b

)
2

y
2

]

3

/

2

k
q
y

[

x
2

y
2

]

3

/

2

k
Q
y

[

(
x
−
b

)
2

y
2

]

3

/

2

]

y
.

{\displaystyle \vec{E\_P} = \left[\frac{kQ(x+b)}{\left[(x+b)^2 + y^2 \right]^{3/2}} + \frac{kqx}{\left[x^2 + y^2 \right]^{3/2}} + \frac{kQ(x-b)}{\left[(x-b)^2 + y^2 \right]^{3/2}}\right]\boldsymbol{i} + \left[\frac{kQy}{\left[(x+b)^2 + y^2 \right]^{3/2}} + \frac{kqy}{\left[x^2 + y^2 \right]^{3/2}} + \frac{kQy}{\left[(x-b)^2 + y^2 \right]^{3/2}}\right]\boldsymbol{y}.}

**Part 2**

Move charge

q

{\displaystyle +q }

 to a point

(

x

0

,
0
)

{\displaystyle (x\_{0}, 0)}

 and then release. The force field vectors are drawn when

q

{\displaystyle +q }

 is at a general position

(
x
,
0
)

{\displaystyle (x,0) }

. Thus, the net Coulomb’s Force on this charge is given by:

F

n
e
t

→

=

F
1

→

F
2

→

{\displaystyle \vec{F\_{net}} = \vec{F\_1} + \vec{F\_2}}

where

F
1

→

=

k
Q
q

r
1
2

r
1

→

r
1

=

k
Q
q

(
x
b

)
2

i

F
2

→

=
−

k
Q
q

r
2
2

r
2

→

r
2

=
−

k
Q
q

(
x
−
b

)
2

i
.

{\displaystyle
\begin{align}
\vec{F\_1} &=\frac{kQq}{r^2\_1} \frac{\vec{r\_1}}{|r\_1|} = \frac{kQq}{(x+b)^2}\boldsymbol{i}\\[5pt]
\vec{F\_2} &=-\frac{kQq}{r^2\_2} \frac{\vec{r\_2}}{|r\_2|} = -\frac{kQq}{(x-b)^2}\boldsymbol{i}.
\end{align} }

Consequently,

F

n
e
t

→

=
k
Q
q

[

1

(
x
b

)
2

−

1

(
x
−
b

)
2

]

i
.

{\displaystyle \vec{F\_{net}} = kQq\left[\frac{1}{(x+b)^2}-\frac{1}{(x-b)^2} \right]\boldsymbol{i}.}

Rewrite the net force as

F

n
e
t

→

=
k
Q
q

[

1

b
2

(

1

x
b

)

2

−

1

b
2

(

1
−

x
b

)

2

]

i

=

k
Q
q

b
2

[

(

1

x
b

)

−
2

−

(

1
−

x
b

)

−
2

]

i
.

{\displaystyle \begin{align}
\vec{F\_{net}} &= kQq\left[\frac{1}{b^2 \left(1+\frac{x}{b}\right)^2}-\frac{1}{b^2 \left(1-\frac{x}{b}\right)^2} \right]\boldsymbol{i} \\[5 pt]
&= \frac{kQq}{b^2}\left[\left(1+\frac{x}{b}\right)^{-2}-\left(1-\frac{x}{b}\right)^{-2} \right]\boldsymbol{i}.
\end{align}}

If

x
<<
b

{\displaystyle x<<b }

, we can use the binomial expansions in the hint:

F

n
e
t

→

=

k
Q
q

b
2

[

(

1
−
2

x
b

)

−

(

1
2

x
b

)

]

i

=

−
4
k
Q
q

b
3

x
i
.

{\displaystyle \begin{align}
\vec{F\_{net}} &= \frac{kQq}{b^2}\left[\left(1-2\frac{x}{b}\right)-\left(1+2\frac{x}{b}\right) \right]\boldsymbol{i} \\[5 pt]
&= \frac{-4kQq}{b^3}x\boldsymbol{i}.
\end{align}}

Using Newton's second law of motion, we get

F

n
e
t

→

=
m
a

−
4
k
Q
q

b
3

x

=
m

d
2

x

d

t
2

.

{\displaystyle \begin{align}
\vec{F\_{net}} &= ma \\[5 pt]
\frac{-4kQq}{b^3}x &= m \frac{d^2 x}{dt^2}.
\end{align}}

Rearrange to get the differential equation in the form of a harmonic oscillator

d
2

x

d

t
2

4
k
Q
q

m

b
3

# x

{\displaystyle \frac{d^2 x}{dt^2} + \frac{4kQq}{mb^3}x = 0.
}

**Part 3**

The differential equation of the harmonic oscillator is

d
2

x

d

t
2

ω
2

# x
0

{\displaystyle \frac{d^2x}{dt^2}+\omega^2x=0}

where

ω

{\displaystyle \omega}

 is the angular frequency. Consequently, the angular frequency is

ω
2

=

4
k
Q
q

m

b
3

ω

=

4
k
Q
q

m

b
3

.

{\displaystyle \begin{align}
\omega^2 &= \frac{4kQq}{mb^3} \\[5 pt]
\omega &= \sqrt{\frac{4kQq}{mb^3}}.
\end{align}
}

The period of oscillation is

T

=

2
π

ω

T

=
2
π

m

b
3

4
k
Q
q

.

{\displaystyle \begin{align}
T &= \frac{2\pi}{\omega}\\[5 pt]
T &= 2\pi \sqrt{\frac{mb^3}{4kQq}}.
\end{align}
}