A uniform electric field `400N/C` acts along positive y-axis, P is a point having co-ordinates (0.6m,-0.2m) and R another point, with co-ordinates (-0.4m, 0.6m). If electric potential at P is 200V, potential at R will be.

Options

A. 100
B. `-150V`
C. 80V
D. `-120V`

Explanation

To find the electric potential at point R given the electric potential at point P and the uniform electric field, we can follow these steps:

### Step 1: Understand the electric field and its direction
The electric field \( E \) is given as \( 400 \, \text{N/C} \) acting along the positive y-axis. This means that the electric field vector can be represented as:
\[ \vec{E} = 400 \hat{j} \, \text{N/C} \]

### Step 2: Identify the coordinates of points P and R
The coordinates of point P are \( (0.6 \, \text{m}, -0.2 \, \text{m}) \) and the coordinates of point R are \( (-0.4 \, \text{m}, 0.6 \, \text{m}) \).

### Step 3: Calculate the change in the y-coordinate
To find the change in potential between points P and R, we need to calculate the change in the y-coordinate:
- The y-coordinate of point P is \( -0.2 \, \text{m} \).
- The y-coordinate of point R is \( 0.6 \, \text{m} \).

The change in y-coordinate (\( \Delta y \)) is:
\[ \Delta y = y_R - y_P = 0.6 - (-0.2) = 0.6 + 0.2 = 0.8 \, \text{m} \]

### Step 4: Use the formula for potential difference
The potential difference \( \Delta V \) between two points in an electric field is given by:
\[ \Delta V = -E \Delta y \]
Here, \( E = 400 \, \text{N/C} \) and \( \Delta y = 0.8 \, \text{m} \).

Substituting the values:
\[ \Delta V = -400 \times 0.8 = -320 \, \text{V} \]

### Step 5: Relate the potentials at points P and R
We know that:
\[ V_R - V_P = \Delta V \]
Substituting the known values:
\[ V_R - 200 = -320 \]

### Step 6: Solve for the potential at point R
Rearranging the equation to find \( V_R \):
\[ V_R = -320 + 200 = -120 \, \text{V} \]

### Final Answer
The electric potential at point R is:
\[ \boxed{-120 \, \text{V}} \]