Gravitational field in `x-y` plane is given as
`E = (2xhati+3y^(2)hatj)N//kg`
Find difference in gravitation potential between two points A and B, where
co-ordinates of A and B are `(2m, 4m) and (6m,0)`.

Explanation

To find the difference in gravitational potential between two points A and B in the given gravitational field, we can follow these steps:

### Step 1: Understand the Given Gravitational Field
The gravitational field is given as:
\[ \mathbf{E} = (2x \hat{i} + 3y^2 \hat{j}) \, \text{N/kg} \]
This means that the gravitational field has a component in the x-direction and a component in the y-direction.

### Step 2: Use the Relationship Between Gravitational Field and Potential
The relationship between the gravitational field \(\mathbf{E}\) and gravitational potential \(V\) is given by:
\[ \mathbf{E} = -\nabla V \]
or
\[ dV = -\mathbf{E} \cdot d\mathbf{r} \]

### Step 3: Define the Differential Displacement
The differential displacement \(d\mathbf{r}\) can be expressed as:
\[ d\mathbf{r} = dx \hat{i} + dy \hat{j} \]

### Step 4: Calculate the Dot Product
Now, we can calculate the dot product:
\[ dV = -\mathbf{E} \cdot d\mathbf{r} = - (2x \hat{i} + 3y^2 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]
This expands to:
\[ dV = - (2x \, dx + 3y^2 \, dy) \]

### Step 5: Integrate to Find the Potential Difference
To find the potential difference \(V_A - V_B\) between points A and B, we integrate:
\[ V_A - V_B = -\int_A^B (2x \, dx + 3y^2 \, dy) \]

### Step 6: Set the Limits of Integration
The coordinates of points A and B are:
- Point A: \( (2 \, \text{m}, 4 \, \text{m}) \)
- Point B: \( (6 \, \text{m}, 0 \, \text{m}) \)

### Step 7: Perform the Integration
We can perform the integration in two parts, first with respect to \(x\) and then with respect to \(y\).

1. **Integrate with respect to \(x\)** from \(x = 2\) to \(x = 6\) while keeping \(y\) constant at \(y = 4\):
\[ -\int_{2}^{6} 2x \, dx = -\left[ x^2 \right]_{2}^{6} = -\left[ 6^2 - 2^2 \right] = -\left[ 36 - 4 \right] = -32 \]

2. **Integrate with respect to \(y\)** from \(y = 4\) to \(y = 0\) while keeping \(x\) constant at \(x = 6\):
\[ -\int_{4}^{0} 3y^2 \, dy = -\left[ y^3 \right]_{4}^{0} = -\left[ 0 - 4^3 \right] = 64 \]

### Step 8: Combine the Results
Now, we combine the results of both integrations:
\[ V_A - V_B = -32 + 64 = 32 \, \text{J/kg} \]

### Final Answer
Thus, the difference in gravitational potential between points A and B is:
\[ V_A - V_B = 32 \, \text{J/kg} \]

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