Electric Dipole

## **Problem**

An electric dipole is placed in the uniform electric field

E
→

=
E
j

{\displaystyle \vec{E} = E \boldsymbol j }

 (due to parallel charged plates) as shown. The charge on the left, located at

(

L
2

,

d
3

)

{\displaystyle \left(\frac{L}{2}, \frac{d}{3}\right)}

, is

−
2
q

{\displaystyle -2q}

; the charge on the right, located at

(

2
L

3

,

2
d

3

)

{\displaystyle \left(\frac{2L}{3}, \frac{2d}{3}\right)}

, is

2
q

{\displaystyle +2q}

.

**Part 1:** Find the initial torque acting on the dipole using the mechanical method.

**Part 2:** Find the initial torque acting on the dipole using the electrical method.

**Part 3:** Write an expression for the electric dipole moment vector when the dipole is in alignment position with the external electric field.

## **Solution**

**Part 1**

Let the origin

(
0
,
0
)

{\displaystyle (0,0) }

 be the pivot point. The net torque is

τ

n
e
t

=

r
1

→

×

F
1

→

r
2

→

×

F
2

→

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

 where the lever arms are

r
1

→

=

L
2

i

d
3

j

{\displaystyle \vec{r\_1} = \frac{L}{2}\boldsymbol{i} + \frac{d}{3}\boldsymbol{j}}

 and

r
1

→

=

2
L

3

i

2
d

3

j

{\displaystyle \vec{r\_1} = \frac{2L}{3}\boldsymbol{i} + \frac{2d}{3}\boldsymbol{j}}

. The forces are

F
1

→

=
−
2
q
E
j

{\displaystyle \vec{F\_1} = -2qE\boldsymbol{j}}

 and

F
2

→

=
2
q
E
j

{\displaystyle \vec{F\_2} = 2qE\boldsymbol{j}}

.

Calculate each cross-product:

r
1

→

×

F
1

→

=

i

j

k

L
2

d
3

0

0

−
2
q
E

0

=
i

d
3

0

−
2
q
E

0

−
j

L
2

0

0

0

k

L
2

d
3

0

−
2
q
E

=
i

[

(

d
3

⋅
0

)

−

(

0
⋅
−
2
q
E

)

]

−
j

[

(

L
2

⋅
0

)

−

(

0
⋅
0

)

]

k

[

(

L
2

⋅
−
2
q
E

)

−

(

d
3

⋅
0

)

]

=
−
q
E
L
k

{\displaystyle \begin{align}
\vec{r\_1} \times \vec{F\_1} &= \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ \frac{L}{2} & \frac{d}{3} & 0 \\ 0 & -2qE & 0 \end{vmatrix} \\[5 pt]
&= \boldsymbol{i}\begin{vmatrix} \frac{d}{3} & 0 \\ -2qE & 0 \end{vmatrix} - \boldsymbol{j} \begin{vmatrix} \frac{L}{2} & 0 \\ 0 & 0 \end{vmatrix} + \boldsymbol{k}\begin{vmatrix} \frac{L}{2} & \frac{d}{3} \\ 0 & -2qE \end{vmatrix} \\[5 pt]
&= \boldsymbol{i}\left[\left(\frac{d}{3} \cdot 0\right) - \left(0 \cdot -2qE\right)\right] - \boldsymbol{j}\left[\left(\frac{L}{2} \cdot 0 \right) - \left(0 \cdot 0 \right)\right] + \boldsymbol{k}\left[\left(\frac{L}{2} \cdot -2qE \right) - \left(\frac{d}{3} \cdot 0\right)\right] \\[5 pt]
&= -qEL \boldsymbol{k}
\end{align}}

r
2

→

×

F
2

→

=

i

j

k

2
L

3

2
d

3

0

0

2
q
E

0

=
i

2
d

3

0

2
q
E

0

−
j

2
L

3

0

0

0

k

2
L

3

2
d

3

0

2
q
E

=
i

[

(

2
d

3

⋅
0

)

−

(

0
⋅
2
q
E

)

]

−
j

[

(

2
L

3

⋅
0

)

−

(

0
⋅
0

)

]

k

[

(

2
L

3

⋅
2
q
E

)

−

(

2
d

3

⋅
0

)

]

=

4
q
E
L

3

k

{\displaystyle \begin{align}
\vec{r\_2} \times \vec{F\_2} &= \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ \frac{2L}{3} & \frac{2d}{3} & 0 \\ 0 & 2qE & 0 \end{vmatrix} \\[5 pt]
&= \boldsymbol{i}\begin{vmatrix} \frac{2d}{3} & 0 \\ 2qE & 0 \end{vmatrix} - \boldsymbol{j} \begin{vmatrix} \frac{2L}{3} & 0 \\ 0 & 0 \end{vmatrix} + \boldsymbol{k}\begin{vmatrix} \frac{2L}{3} & \frac{2d}{3} \\ 0 & 2qE \end{vmatrix} \\[5 pt]
&= \boldsymbol{i}\left[\left(\frac{2d}{3} \cdot 0\right) - \left(0 \cdot 2qE\right)\right] - \boldsymbol{j}\left[\left(\frac{2L}{3} \cdot 0 \right) - \left(0 \cdot 0 \right)\right] + \boldsymbol{k}\left[\left(\frac{2L}{3} \cdot 2qE \right) - \left(\frac{2d}{3} \cdot 0\right)\right] \\[5 pt]
&= \frac{4qEL}{3} \boldsymbol{k}
\end{align}}

Adding the two torques yields:

τ

n
e
t

=
−
q
E
L
k

4
q
E
L

3

k

=

q
E
L

3

k
.

{\displaystyle \begin{align}
{\tau}\_{net} &= -qEL\boldsymbol{k} + \frac{4qEL}{3}\boldsymbol{k}\\[5 pt]
&= \frac{qEL}{3}\boldsymbol{k}.
\end{align}
}

**Part 2**

The net torque of an electric dipole can also be computed using the formula

τ

n
e
t

=

p
→

×

E
→

{\displaystyle {\tau}\_{net} = \vec{p} \times \vec{E} }

.

The initial dipole moment vector is given by:

p
→

=
2
q

[

(

2
L

3

−

L
2

)

i

(

2
d

3

−

d
3

)

j

]

=

q
L

3

i

2
q
d

3

j

{\displaystyle \begin{align}
\vec{p} &= 2q\left[\left(\frac{2L}{3} - \frac{L}{2} \right)\boldsymbol{i} + \left(\frac{2d}{3} - \frac{d}{3} \right)\boldsymbol{j} \right]\\[5 pt]
&= \frac{qL}{3}\boldsymbol{i} + \frac{2qd}{3} \boldsymbol{j}
\end{align}
}

Thus,

p
→

×

E
→

=

i

j

k

q
L

3

2
q
d

3

0

0

E

0

=
i

2
q
d

3

0

E

0

−
j

q
L

3

0

0

0

k

q
L

3

2
q
d

3

0

E

=
i

[

(

2
q
d

3

⋅
0

)

−

(

0
⋅
0

)

]

−
j

[

(

q
L

3

⋅
0

)

−

(

0
⋅
0

)

]

k

[

(

q
L

3

⋅
E

)

−

(

2
q
d

3

⋅
0

)

]

=

q
L
E

3

k

{\displaystyle \begin{align}
\vec{p} \times \vec{E} &= \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ \frac{qL}{3} & \frac{2qd}{3} & 0 \\ 0 & E & 0 \end{vmatrix} \\[5 pt]
&= \boldsymbol{i}\begin{vmatrix} \frac{2qd}{3} & 0 \\ E & 0 \end{vmatrix} - \boldsymbol{j} \begin{vmatrix} \frac{qL}{3} & 0 \\ 0 & 0 \end{vmatrix} + \boldsymbol{k}\begin{vmatrix} \frac{qL}{3} & \frac{2qd}{3} \\ 0 & E \end{vmatrix} \\[5 pt]
&= \boldsymbol{i}\left[\left(\frac{2qd}{3} \cdot 0\right) - \left(0 \cdot 0\right)\right] - \boldsymbol{j}\left[\left(\frac{qL}{3} \cdot 0 \right) - \left(0 \cdot 0 \right)\right] + \boldsymbol{k}\left[\left(\frac{qL}{3} \cdot E \right) - \left(\frac{2qd}{3} \cdot 0\right)\right] \\[5 pt]
&= \frac{qLE}{3} \boldsymbol{k}
\end{align}}

which agrees with the answer in Part 1.

**Part 3**

The dipole will rotate counterclockwise and at some point the dipole moment vector will align with the electric field. The length of the dipole moment vector is constant and it is equal to

p
→

=
q

(

L
3

)

2

(

2
d

3

)

2

.

{\displaystyle \begin{align}
\end{align}
}

The vector

p
→

{\displaystyle \vec{p}}

 will point in the same direction of the electric field:

p
→

=
q

(

L
3

)

2

(

2
d

3

)

2

j
.

{\displaystyle \begin{align}
\vec{p} = q \sqrt{\left(\frac{L}{3}\right)^2 + \left(\frac{2d}{3}\right)^2} \boldsymbol{j}.
\end{align}
}