A constant electric field of intensity 36N/C exists along the z-axis. If P and Q be two points whose coordinates are (10 cm, 0, -20 cm) and (0, -10 cm, 30 cm) respectively, then find the potential difference `V_(P)-V_(Q)`.

Answer

To find the potential difference \( V_P - V_Q \) between the points P and Q in a constant electric field, we can follow these steps:

### Step 1: Identify the electric field and its direction
The electric field intensity is given as \( E = 36 \, \text{N/C} \) along the z-axis. This means that the electric field vector can be represented as:
\[ \vec{E} = 36 \, \hat{k} \, \text{N/C} \]

### Step 2: Identify the coordinates of points P and Q
The coordinates of the points are:
- Point P: \( (10 \, \text{cm}, 0, -20 \, \text{cm}) \) or \( (0.1 \, \text{m}, 0, -0.2 \, \text{m}) \)
- Point Q: \( (0, -10 \, \text{cm}, 30 \, \text{cm}) \) or \( (0, -0.1 \, \text{m}, 0.3 \, \text{m}) \)

### Step 3: Calculate the potential difference using the formula
The potential difference \( V_P - V_Q \) in an electric field can be calculated using the formula:
\[ V_P - V_Q = -\int_{Q}^{P} \vec{E} \cdot d\vec{r} \]
Since the electric field is constant and directed along the z-axis, we can simplify the integral.

### Step 4: Determine the change in the z-coordinate
The z-coordinates for points P and Q are:
- For P: \( z_P = -20 \, \text{cm} = -0.2 \, \text{m} \)
- For Q: \( z_Q = 30 \, \text{cm} = 0.3 \, \text{m} \)

The change in z-coordinate as we move from Q to P is:
\[ z_P - z_Q = -0.2 - 0.3 = -0.5 \, \text{m} \]

### Step 5: Substitute into the potential difference formula
Substituting the values into the potential difference formula gives:
\[ V_P - V_Q = -\int_{Q}^{P} \vec{E} \cdot d\vec{r} = -E \cdot (z_P - z_Q) \]
Substituting \( E = 36 \, \text{N/C} \):
\[ V_P - V_Q = -36 \cdot (-0.5) = 18 \, \text{V} \]

### Final Answer
Thus, the potential difference \( V_P - V_Q \) is:
\[ V_P - V_Q = 18 \, \text{V} \]