An electric charge `10^(-8) C` is placed at the point (4m, 7m, 2m). At the point (1m, 3m, 2m), the electric :

Options

A. potential will be 18 V
B. field has no Z-component
C. field will be along Z-axis
D. potential will be 1.8 V

Explanation

To solve the problem step by step, we need to find the electric potential and electric field at the point (1m, 3m, 2m) due to a charge of \(10^{-8} C\) located at (4m, 7m, 2m).

### Step 1: Identify the given values
- Charge, \( Q = 10^{-8} C \)
- Position of charge, \( P_1 = (4m, 7m, 2m) \)
- Point of interest, \( P_2 = (1m, 3m, 2m) \)

### Step 2: Calculate the displacement vector \( \vec{r} \)
The displacement vector \( \vec{r} \) from the charge to the point of interest can be calculated as:
\[ \vec{r} = P_2 - P_1 = (1 - 4) \hat{i} + (3 - 7) \hat{j} + (2 - 2) \hat{k} \]
\[ \vec{r} = -3 \hat{i} - 4 \hat{j} + 0 \hat{k} \]

### Step 3: Calculate the magnitude of \( \vec{r} \)
The magnitude of the displacement vector \( r \) is given by:
\[ r = |\vec{r}| = \sqrt{(-3)^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, m \]

### Step 4: Calculate the electric potential \( V \)
The electric potential \( V \) at point \( P_2 \) due to the charge \( Q \) is given by:
\[ V = \frac{k \cdot Q}{r} \]
Where \( k \) (Coulomb's constant) is approximately \( 9 \times 10^9 \, N \cdot m^2/C^2 \).
Substituting the values:
\[ V = \frac{9 \times 10^9 \cdot 10^{-8}}{5} = \frac{9 \times 10^1}{5} = 18 \, V \]

### Step 5: Calculate the electric field \( \vec{E} \)
The electric field \( \vec{E} \) at point \( P_2 \) due to the charge \( Q \) can be calculated using the formula:
\[ \vec{E} = \frac{k \cdot Q}{r^2} \cdot \hat{r} \]
Where \( \hat{r} \) is the unit vector in the direction of \( \vec{r} \):
\[ \hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{-3 \hat{i} - 4 \hat{j}}{5} = -\frac{3}{5} \hat{i} - \frac{4}{5} \hat{j} \]
Now substituting into the electric field equation:
\[ \vec{E} = \frac{9 \times 10^9 \cdot 10^{-8}}{5^2} \cdot \left(-\frac{3}{5} \hat{i} - \frac{4}{5} \hat{j}\right) \]
\[ = \frac{9 \times 10^1}{25} \cdot \left(-\frac{3}{5} \hat{i} - \frac{4}{5} \hat{j}\right) \]
\[ = \frac{18}{25} \left(-3 \hat{i} - 4 \hat{j}\right) \]

### Final Results
- The electric potential \( V \) at point \( (1m, 3m, 2m) \) is \( 18 \, V \).
- The electric field \( \vec{E} \) is given by:
\[ \vec{E} = \frac{18}{25} \left(-3 \hat{i} - 4 \hat{j}\right) \]