A uniform electric field `E_0` is directed along positive y-driection. Find the change in electric potential energy of a positive test charge `q_0` when it is displaced in this field from `y_i=a` to `y_f=2a` along the y-axis.

Explanation

To find the change in electric potential energy of a positive test charge \( q_0 \) when it is displaced in a uniform electric field \( E_0 \) from \( y_i = a \) to \( y_f = 2a \) along the y-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Electric Field and Electric Potential:**
The electric potential \( V \) is related to the electric field \( E \) by the equation:
\[ V = -\int E \, dy \]
For a uniform electric field \( E_0 \) directed along the positive y-direction, the potential difference \( \Delta V \) between two points \( y_i \) and \( y_f \) is given by:
\[ \Delta V = -E_0 (y_f - y_i) \]

2. **Calculate the Change in Potential:**
Here, \( y_i = a \) and \( y_f = 2a \). Thus, the change in potential \( \Delta V \) can be calculated as:
\[ \Delta V = -E_0 (2a - a) = -E_0 (a) = -E_0 a \]

3. **Relate Change in Potential to Change in Potential Energy:**
The change in electric potential energy \( \Delta U \) of the charge \( q_0 \) is given by:
\[ \Delta U = q_0 \Delta V \]
Substituting the expression for \( \Delta V \):
\[ \Delta U = q_0 (-E_0 a) = -q_0 E_0 a \]

4. **Interpret the Result:**
The negative sign indicates that the electric potential energy decreases as the charge moves in the direction of the electric field.

### Final Answer:
The change in electric potential energy of the positive test charge \( q_0 \) when it is displaced from \( y_i = a \) to \( y_f = 2a \) is:
\[ \Delta U = -q_0 E_0 a \]