A charged particle having charge `q` and mass `m` is projected into a region of uniform electric field of strength `vecE_(0)`, with velocity `vecV_(0)` perpendicular to `vecE_(0)`. Throughout the motion apart from electric force particle also experiences a dissipatice force of constant magnitude `qE_(0)` and directed opposite to its velocity. If `|vecV_(0)|=6m//s`, then find its speed when it has turned through an angle of `90^(@)`.

Explanation

To solve the problem, we will analyze the motion of the charged particle in the presence of a uniform electric field and a dissipative force.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The charged particle experiences two forces:
- The electric force due to the electric field \( \vec{E}_0 \): \( \vec{F}_e = q \vec{E}_0 \).
- The dissipative force, which is constant in magnitude and opposite to the velocity: \( \vec{F}_d = -q E_0 \hat{v} \), where \( \hat{v} \) is the unit vector in the direction of velocity.
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