An electric field of 1000 V/m is applied to an electric dipole at angle of 45 degrees. The value of

Q: Why is there a negative sign in the potential energy formula?

The negative sign arises from the definition of potential energy as the negative of the work done by conservative forces. When a dipole aligns with the electric field, the field does positive work, so the potential energy must decrease. It reaches its minimum (most negative) value when perfectly aligned.

Q: What does zero potential energy mean for a dipole?

By convention, we set the potential energy to zero when the dipole is perpendicular to the electric field ($\theta = 90^\circ$). This is an arbitrary reference point, much like choosing sea level as zero gravitational potential energy.

Q: When is the dipole in stable equilibrium?

Stable equilibrium occurs when the potential energy is at its absolute minimum. For a dipole, this happens when it is perfectly aligned with the electric field ($\theta = 0^\circ$), giving $U = -pE$.

Q: Can I use $\sin\theta$ instead of $\cos\theta$ in the formula?

No. The potential energy is derived from a dot product ($\mathbf{p} \cdot \mathbf{E}$), which inherently uses the cosine of the angle between the vectors. Using sine would give you the magnitude of the torque ($\tau = pE \sin\theta$), not the energy.

Q: Do I need to convert any units before calculating?

Always check your units! In this problem, the electric field is in V/m (which is equivalent to N/C) and the dipole moment is in C-m. Both are standard SI units, so you can multiply them directly to get the energy in Joules.

Question

An electric field of 1000 V/m is applied to an electric dipole at angle of $45^\circ$. The value of electric dipole moment is $10^{-29}$ C-m. What is the potential energy of the electric dipole?

Options:

(A) $-9 	imes 10^{-20}$ J

(B) $-10 	imes 10^{-29}$ J

(C) $-20 	imes 10^{-18}$ J

(D) $-7 	imes 10^{-27}$ J

Sigma Percentile · Accepted Answer

### The Dance of the Dipole

Imagine a tiny compass needle, but instead of pointing north, it's an electric dipole trying to align itself with an invisible river of force—a uniform electric field. This problem is a classic exploration of the energy dynamics when such a dipole is placed at an angle to the field.

When an electric dipole is placed in a uniform electric field, it experiences a torque that tries to rotate it so that its dipole moment vector $\mathbf{p}$ aligns perfectly with the electric field vector $\mathbf{E}$. Because the field exerts a force to cause this rotation, the dipole possesses potential energy based on its orientation.

### The Master Equation

The potential energy $U$ of an electric dipole in a uniform electric field is elegantly captured by the negative dot product of the dipole moment and the electric field vectors:

$$U = -\mathbf{p} \cdot \mathbf{E}$$

Expanding this dot product, we get the scalar form:

$$U = -pE \cos	heta$$

Here, $	heta$ is the angle between the dipole moment and the electric field. The negative sign is profoundly important. It tells us that the potential energy is minimum (most stable) when the dipole is perfectly aligned with the field ($	heta = 0^\circ$, $\cos 0^\circ = 1$, $U = -pE$). Conversely, the energy is maximum (most unstable) when it's pointing exactly opposite to the field ($	heta = 180^\circ$, $\cos 180^\circ = -1$, $U = +pE$).

### Executing the Calculation

In our specific scenario, we are given:
- Electric field strength, $E = 1000 	ext{ V/m} = 10^3 	ext{ V/m}$
- Dipole moment, $p = 10^{-29} 	ext{ C-m}$
- Angle, $	heta = 45^\circ$

Let's substitute these values into our master equation:

$$U = -(10^{-29}) 	imes (10^3) 	imes \cos(45^\circ)$$

We know that $\cos(45^\circ) = \frac{1}{\sqrt{2}}$. Substituting this in, we get:

$$U = -10^{-26} 	imes \frac{1}{\sqrt{2}}$$

Now, we use the standard approximation $\frac{1}{\sqrt{2}} \approx 0.707$:

$$U \approx -0.707 	imes 10^{-26} 	ext{ J}$$

To match the format of our multiple-choice options, we adjust the scientific notation by moving the decimal point one place to the right, which decreases the exponent by one:

$$U = -7.07 	imes 10^{-27} 	ext{ J}$$

Looking at the given options, this value is approximately **$-7 	imes 10^{-27} 	ext{ J}$**. The beauty of this problem lies in its direct application of a fundamental concept, reminding us that even the smallest particles obey the grand laws of energy and alignment.

Animated Solution for Physics - Electrostatics: An electric field of 1000 V/m is applied to an electric dipole at angle of $4 5^{\circ}$ . The value of electric dipole moment is $1 0^{- 29}$ C-m. What is the potential energy of the electric dipole?

Select Answer:

Visualized Solution

$Visualizing the Setup$

$The Master Equation$

$Substituting the Values$

$Mathematical Execution$

$Final Answer$

The Sigma Insight: Electric Dipole

The Dance of the Dipole

The Master Equation

Executing the Calculation

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