Electric dipole of moment `vecp = phati` is kept at a point (x,y) in an electric field `vecE = 4xy^2hati + 4x^2yhatj`. Find the force on the dipole.

Options

A. `4py^2`
B. `4py(y^2 + 2x^2)^1//2`
C. `4py^2 + 8pxy`
D. `4py(y^2 + 4x^2)^1//2`

Explanation

To find the force on an electric dipole in an electric field, we can use the formula for the force on a dipole moment \(\vec{p}\) in an electric field \(\vec{E}\):

\[ \vec{F} = \nabla (\vec{p} \cdot \vec{E}) \]

Given:
- The electric dipole moment \(\vec{p} = p \hat{i}\)
- The electric field \(\vec{E} = 4xy^2 \hat{i} + 4x^2y \hat{j}\)

### Step 1: Calculate the dot product \(\vec{p} \cdot \vec{E}\)

\[ \vec{p} \cdot \vec{E} = (p \hat{i}) \cdot (4xy^2 \hat{i} + 4x^2y \hat{j}) = p(4xy^2) + 0 = 4pxy^2 \]

### Step 2: Find the gradient of \(\vec{p} \cdot \vec{E}\)

We need to compute the gradient \(\nabla(4pxy^2)\). In Cartesian coordinates, the gradient operator is given by:

\[ \nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} \]

Calculating the partial derivatives:

- \(\frac{\partial}{\partial x}(4pxy^2) = 4py^2\)
- \(\frac{\partial}{\partial y}(4pxy^2) = 8pxy\)

Thus, the gradient is:

\[ \nabla(4pxy^2) = \hat{i}(4py^2) + \hat{j}(8pxy) \]

### Step 3: Write the force vector \(\vec{F}\)

Substituting the gradient into the force equation:

\[ \vec{F} = \nabla(4pxy^2) = 4py^2 \hat{i} + 8pxy \hat{j} \]

### Final Answer

The force on the dipole is:

\[ \vec{F} = 4py^2 \hat{i} + 8pxy \hat{j} \]

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