Two point charges q and –q are at positions (0,0,d) and (0,0, –d) respectively . What is the electric field at (a,0,0 ) ?

Options

A. `(2qd)/(4piepsilon_0(d^2+a^2)^(3//2))hatk`
B. `(qd)/(4piepsilon_0(d^2+a^2)^(3//2))hatk`
C. `(-2qd)/(4piepsilon_0(d^2+a^2)^(3//2))hatk`
D. `(-qd)/(4piepsilon_0(d^2+a^2)^(3//2))hatk`

Answer

To find the electric field at the point (a, 0, 0) due to two point charges \( q \) and \( -q \) located at (0, 0, d) and (0, 0, -d) respectively, we can follow these steps:

### Step 1: Identify the positions of the charges
- Charge \( q \) is located at \( (0, 0, d) \).
- Charge \( -q \) is located at \( (0, 0, -d) \).
- We need to find the electric field at the point \( (a, 0, 0) \).

### Step 2: Calculate the distance from each charge to the point (a, 0, 0)
- The distance \( r_1 \) from charge \( q \) at \( (0, 0, d) \) to the point \( (a, 0, 0) \) is given by:
\[ r_1 = \sqrt{(a - 0)^2 + (0 - 0)^2 + (0 - d)^2} = \sqrt{a^2 + d^2} \]
- The distance \( r_2 \) from charge \( -q \) at \( (0, 0, -d) \) to the point \( (a, 0, 0) \) is given by:
\[ r_2 = \sqrt{(a - 0)^2 + (0 - 0)^2 + (0 + d)^2} = \sqrt{a^2 + d^2} \]

### Step 3: Calculate the electric field due to each charge
- The electric field \( E_1 \) due to charge \( q \) at the point \( (a, 0, 0) \) is directed away from the charge (since it is positive):
\[ E_1 = \frac{kq}{r_1^2} = \frac{kq}{a^2 + d^2} \]
The direction of \( E_1 \) is along the vector pointing from \( (0, 0, d) \) to \( (a, 0, 0) \), which can be expressed as:
\[ \hat{r_1} = \frac{(a, 0, 0) - (0, 0, d)}{\sqrt{a^2 + d^2}} = \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{-d}{\sqrt{a^2 + d^2}}\right) \]
Thus,
\[ \vec{E_1} = E_1 \hat{r_1} = \frac{kq}{a^2 + d^2} \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{-d}{\sqrt{a^2 + d^2}}\right) \]

- The electric field \( E_2 \) due to charge \( -q \) at the point \( (a, 0, 0) \) is directed towards the charge (since it is negative):
\[ E_2 = \frac{k(-q)}{r_2^2} = -\frac{kq}{a^2 + d^2} \]
The direction of \( E_2 \) is towards the charge at \( (0, 0, -d) \):
\[ \hat{r_2} = \frac{(a, 0, 0) - (0, 0, -d)}{\sqrt{a^2 + d^2}} = \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{d}{\sqrt{a^2 + d^2}}\right) \]
Thus,
\[ \vec{E_2} = E_2 \hat{r_2} = -\frac{kq}{a^2 + d^2} \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{d}{\sqrt{a^2 + d^2}}\right) \]

### Step 4: Combine the electric fields
- The total electric field \( \vec{E}_{\text{net}} \) at point \( (a, 0, 0) \) is the vector sum of \( \vec{E_1} \) and \( \vec{E_2} \):
\[ \vec{E}_{\text{net}} = \vec{E_1} + \vec{E_2} \]
\[ \vec{E}_{\text{net}} = \frac{kq}{a^2 + d^2} \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{-d}{\sqrt{a^2 + d^2}}\right) - \frac{kq}{a^2 + d^2} \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{d}{\sqrt{a^2 + d^2}}\right) \]
\[ = \frac{kq}{a^2 + d^2} \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{-d - d}{\sqrt{a^2 + d^2}}\right) \]
\[ = \frac{kq}{a^2 + d^2} \left(\frac{a}{\sqrt{a^2 + d^2}}, 0, \frac{-2d}{\sqrt{a^2 + d^2}}\right) \]

### Step 5: Final expression for the electric field
- Therefore, the electric field at the point \( (a, 0, 0) \) is:
\[ \vec{E}_{\text{net}} = \frac{kq}{(a^2 + d^2)^{3/2}} \left(a, 0, -2d\right) \]