The Electric field is given by `vec( E)= (vec( F))/(q_(0))`, here the test charge `.q_(0).` should be
(a) Infinitesimally small and positive
(b) Infinitesimally small and negative

Options

A. only a
B. only .b.
C. a (or )b
D. neither .a. or .b.

Answer

To solve the question regarding the nature of the test charge \( q_0 \) in the expression for the electric field \( \vec{E} = \frac{\vec{F}}{q_0} \), we can follow these steps:

### Step 1: Understanding the Electric Field
The electric field \( \vec{E} \) at a point in space is defined as the force \( \vec{F} \) experienced by a unit positive charge placed at that point. The formula \( \vec{E} = \frac{\vec{F}}{q_0} \) indicates that the electric field is derived from the force acting on a charge \( q_0 \).

### Step 2: Nature of the Test Charge
To accurately measure the electric field without influencing it, the test charge \( q_0 \) must be:
- **Infinitesimally small**: This ensures that the test charge does not disturb the existing electric field created by other charges.
- **Positive**: By convention, the electric field is defined in terms of the force on a positive test charge. This means that the direction of the electric field is taken to be the direction of the force on a positive charge.

### Step 3: Conclusion
Given the above understanding, the correct choice for the test charge \( q_0 \) is:
- **(a) Infinitesimally small and positive**.

Thus, the answer to the question is that the test charge \( q_0 \) should be infinitesimally small and positive.