At which point is the electric field intensity zero due to an electric dipole?
2 Answers
To find where the electric field intensity is zero due to an electric dipole, let’s consider a classical dipole system with two opposite charges of magnitude \( +q \) and \( -q \) separated by a distance \( 2a \). We’ll analyze the problem in both the axial and equatorial planes of the dipole.
### 1. **Axial Plane**
The axial plane is the line that extends through both charges of the dipole. We will determine the point on this axis where the electric field is zero.
1. **Position of Charges**: Suppose the dipole is placed along the \( x \)-axis with charges \( +q \) and \( -q \) located at \( (a, 0) \) and \( (-a, 0) \) respectively.
2. **Electric Field Expression**:
- The electric field due to the positive charge \( +q \) at a point \( (x, 0) \) on the axis is:
\[
\vec{E}_{+} = \frac{kq}{(x - a)^2} \hat{i}
\]
- The electric field due to the negative charge \( -q \) at the same point \( (x, 0) \) is:
\[
\vec{E}_{-} = -\frac{kq}{(x + a)^2} \hat{i}
\]
3. **Net Electric Field**: The total electric field at \( (x, 0) \) is:
\[
\vec{E} = \vec{E}_{+} + \vec{E}_{-} = \frac{kq}{(x - a)^2} - \frac{kq}{(x + a)^2}
\]
4. **Setting Electric Field to Zero**:
\[
\frac{kq}{(x - a)^2} - \frac{kq}{(x + a)^2} = 0
\]
Simplify to find:
\[
\frac{1}{(x - a)^2} = \frac{1}{(x + a)^2}
\]
Solving this yields no real solution for \( x \) because it implies:
\[
(x - a)^2 = (x + a)^2
\]
which leads to \( -4ax = 0 \), implying that \( x \) must be zero, but that’s not feasible as the field never cancels out in the axial direction directly.
### 2. **Equatorial Plane**
The equatorial plane is perpendicular to the axis of the dipole and equidistant from the charges. Let’s find the point where the electric field is zero in this plane.
1. **Position of the Point**: Consider a point \( (0, y) \) on the equatorial plane.
2. **Electric Field Expression**:
- The electric field due to each charge at this point is:
\[
\vec{E}_{+} = \frac{kq}{(a^2 + y^2)^{3/2}} \left( \frac{a}{\sqrt{a^2 + y^2}} \hat{i} - \frac{y}{\sqrt{a^2 + y^2}} \hat{j} \right)
\]
\[
\vec{E}_{-} = -\frac{kq}{(a^2 + y^2)^{3/2}} \left( \frac{a}{\sqrt{a^2 + y^2}} \hat{i} + \frac{y}{\sqrt{a^2 + y^2}} \hat{j} \right)
\]
3. **Net Electric Field**:
\[
\vec{E} = \vec{E}_{+} + \vec{E}_{-}
\]
The \( \hat{i} \) components cancel out because they are equal and opposite. The \( \hat{j} \) components combine to:
\[
\vec{E}_{y} = \frac{-2kq y}{(a^2 + y^2)^{3/2}} \hat{j}
\]
4. **Setting Electric Field to Zero**:
For \( \vec{E}_{y} \) to be zero:
\[
\frac{-2kq y}{(a^2 + y^2)^{3/2}} = 0
\]
This implies:
\[
y = 0
\]
However, this means the zero field point is actually at infinity along the perpendicular bisector of the dipole for real finite values of \( y \).
### **Conclusion**
The electric field intensity due to a dipole is zero only at infinity along the equatorial plane. In practical terms, a real finite distance from the dipole where the field is exactly zero doesn’t exist, except in idealized theoretical scenarios.
### 1. **Axial Plane**
The axial plane is the line that extends through both charges of the dipole. We will determine the point on this axis where the electric field is zero.
1. **Position of Charges**: Suppose the dipole is placed along the \( x \)-axis with charges \( +q \) and \( -q \) located at \( (a, 0) \) and \( (-a, 0) \) respectively.
2. **Electric Field Expression**:
- The electric field due to the positive charge \( +q \) at a point \( (x, 0) \) on the axis is:
\[
\vec{E}_{+} = \frac{kq}{(x - a)^2} \hat{i}
\]
- The electric field due to the negative charge \( -q \) at the same point \( (x, 0) \) is:
\[
\vec{E}_{-} = -\frac{kq}{(x + a)^2} \hat{i}
\]
3. **Net Electric Field**: The total electric field at \( (x, 0) \) is:
\[
\vec{E} = \vec{E}_{+} + \vec{E}_{-} = \frac{kq}{(x - a)^2} - \frac{kq}{(x + a)^2}
\]
4. **Setting Electric Field to Zero**:
\[
\frac{kq}{(x - a)^2} - \frac{kq}{(x + a)^2} = 0
\]
Simplify to find:
\[
\frac{1}{(x - a)^2} = \frac{1}{(x + a)^2}
\]
Solving this yields no real solution for \( x \) because it implies:
\[
(x - a)^2 = (x + a)^2
\]
which leads to \( -4ax = 0 \), implying that \( x \) must be zero, but that’s not feasible as the field never cancels out in the axial direction directly.
### 2. **Equatorial Plane**
The equatorial plane is perpendicular to the axis of the dipole and equidistant from the charges. Let’s find the point where the electric field is zero in this plane.
1. **Position of the Point**: Consider a point \( (0, y) \) on the equatorial plane.
2. **Electric Field Expression**:
- The electric field due to each charge at this point is:
\[
\vec{E}_{+} = \frac{kq}{(a^2 + y^2)^{3/2}} \left( \frac{a}{\sqrt{a^2 + y^2}} \hat{i} - \frac{y}{\sqrt{a^2 + y^2}} \hat{j} \right)
\]
\[
\vec{E}_{-} = -\frac{kq}{(a^2 + y^2)^{3/2}} \left( \frac{a}{\sqrt{a^2 + y^2}} \hat{i} + \frac{y}{\sqrt{a^2 + y^2}} \hat{j} \right)
\]
3. **Net Electric Field**:
\[
\vec{E} = \vec{E}_{+} + \vec{E}_{-}
\]
The \( \hat{i} \) components cancel out because they are equal and opposite. The \( \hat{j} \) components combine to:
\[
\vec{E}_{y} = \frac{-2kq y}{(a^2 + y^2)^{3/2}} \hat{j}
\]
4. **Setting Electric Field to Zero**:
For \( \vec{E}_{y} \) to be zero:
\[
\frac{-2kq y}{(a^2 + y^2)^{3/2}} = 0
\]
This implies:
\[
y = 0
\]
However, this means the zero field point is actually at infinity along the perpendicular bisector of the dipole for real finite values of \( y \).
### **Conclusion**
The electric field intensity due to a dipole is zero only at infinity along the equatorial plane. In practical terms, a real finite distance from the dipole where the field is exactly zero doesn’t exist, except in idealized theoretical scenarios.
The electric field intensity is zero at a point along the axis perpendicular to the line joining the two charges of the electric dipole. Specifically:
- **For an ideal dipole** (where the separation between the charges is small), the electric field intensity becomes zero along the perpendicular bisector of the dipole. This is the point equidistant from both the positive and negative charges, but only in cases where the dipole is not in a uniform external field.
- **Mathematically**, at any point on the perpendicular bisector, the electric field contributions from the positive and negative charges cancel each other out, leading to zero net electric field intensity.
This applies in the absence of any external fields or other nearby charges.
- **For an ideal dipole** (where the separation between the charges is small), the electric field intensity becomes zero along the perpendicular bisector of the dipole. This is the point equidistant from both the positive and negative charges, but only in cases where the dipole is not in a uniform external field.
- **Mathematically**, at any point on the perpendicular bisector, the electric field contributions from the positive and negative charges cancel each other out, leading to zero net electric field intensity.
This applies in the absence of any external fields or other nearby charges.