A point charge `2 times 10^(-2)` is moved form `P` to `S` in a uniform electric field of `30NC^(-1)` directed along positive `x` -axis.If coordinates of `P` and `S` are `(1,2,0)m` and `(0,0,0)m` respectively,the work done by electric field will be

Options

A. `600mJ`
B. `-1200mJ`
C. `-600mJ`
D. `1200mJ`

Answer

To find the work done by the electric field when moving a point charge from point P to point S, we can follow these steps:

### Step 1: Identify the given values
- Charge, \( Q = 2 \times 10^{-2} \, \text{C} \)
- Electric field, \( E = 30 \, \text{N/C} \)
- Initial position, \( P(1, 2, 0) \, \text{m} \)
- Final position, \( S(0, 0, 0) \, \text{m} \)

### Step 2: Calculate the displacement vector
The displacement vector \( \vec{d} \) from point P to point S can be calculated as:
\[ \vec{d} = \vec{S} - \vec{P} = (0, 0, 0) - (1, 2, 0) = (-1, -2, 0) \, \text{m} \]

### Step 3: Calculate the work done by the electric field
The work done \( W \) by the electric field when moving a charge is given by the formula:
\[ W = Q \cdot \vec{E} \cdot \vec{d} \]
Where \( \vec{E} \) is the electric field vector. Since the electric field is directed along the positive x-axis, we can represent it as:
\[ \vec{E} = (30, 0, 0) \, \text{N/C} \]

### Step 4: Calculate the dot product \( \vec{E} \cdot \vec{d} \)
Now, we calculate the dot product:
\[ \vec{E} \cdot \vec{d} = (30, 0, 0) \cdot (-1, -2, 0) = 30 \times (-1) + 0 \times (-2) + 0 \times 0 = -30 \, \text{N m/C} \]

### Step 5: Substitute into the work formula
Now substituting the values into the work formula:
\[ W = Q \cdot (\vec{E} \cdot \vec{d}) = (2 \times 10^{-2}) \cdot (-30) = -0.6 \, \text{J} \]

### Step 6: Convert to millijoules
To express this in millijoules:
\[ W = -0.6 \, \text{J} = -600 \, \text{mJ} \]

### Final Answer
The work done by the electric field is:
\[ W = -600 \, \text{mJ} \]

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