When a test charge is brough in from infinity along the perpendicular bisector of an electric dipole the work done is

Options

A. positive
B. zero
C. negative
D. none of these

Answer

To solve the question, we need to analyze the situation of bringing a test charge from infinity to the perpendicular bisector of an electric dipole. Here's a step-by-step solution:

### Step 1: Understanding the Electric Dipole
An electric dipole consists of two equal and opposite charges separated by a small distance. The electric field produced by a dipole varies with distance and direction.

**Hint:** Recall the definition and characteristics of an electric dipole, including the dipole moment and the electric field it generates.

### Step 2: Identifying the Perpendicular Bisector
The perpendicular bisector of an electric dipole is a line that is equidistant from both charges of the dipole. At this location, the electric field is directed horizontally, and the potential is constant along this line.

**Hint:** Remember that the perpendicular bisector of a dipole is a line of symmetry where the electric field behaves uniformly.

### Step 3: Understanding Equipotential Lines
The perpendicular bisector of the dipole is an equipotential line, meaning that the electric potential at every point along this line is the same. Therefore, when moving along this line, there is no change in electric potential.

**Hint:** Equipotential lines are defined as lines where the potential is constant. No work is done when moving a charge along these lines.

### Step 4: Work Done in Moving the Charge
The work done \( W \) in moving a charge \( q \) in an electric field is given by the formula:
\[ W = q \Delta V \]
where \( \Delta V \) is the change in electric potential. Since we are moving the test charge along the equipotential line, the change in potential \( \Delta V \) is zero.

**Hint:** Recall that work done is related to the change in potential energy, which depends on the change in electric potential.

### Step 5: Conclusion
Since the change in potential \( \Delta V = 0 \), the work done \( W \) is:
\[ W = q \cdot 0 = 0 \]
Thus, the work done in bringing the test charge from infinity to the perpendicular bisector of the electric dipole is zero.

**Final Answer:** The work done is zero.