A particle of charge `+q` and mass m moving under the influence of a uniform electric field `E hati` and a magnetic field `Bhatk` enters in I quadrant of a coordinate system at a point `(0,a)` with initial velocity `v hati` and leaves the quadrant at a point
`(2a.,0)` with velocity `-2vhatj` Find
(aq) Magnitude of electric field at point `(0,a)` (C ) Rate of work done by both the fields at `(2a,0)` .

Explanation

To solve the problem, we will break it down into two parts as requested:

### Part (a): Magnitude of Electric Field at Point (0, a)

1. **Identify Initial and Final Conditions**:
- The particle enters the first quadrant at point (0, a) with initial velocity \( \vec{v_i} = v \hat{i} \).
- The particle leaves the quadrant at point (2a, 0) with final velocity \( \vec{v_f} = -2v \hat{j} \).

2. **Apply Work-Energy Theorem**:
- The work done by the electric field is equal to the change in kinetic energy of the particle.
- The kinetic energy at the initial point is:
\[ KE_i = \frac{1}{2} m v^2 \]
- The kinetic energy at the final point is:
\[ KE_f = \frac{1}{2} m (-2v)^2 = \frac{1}{2} m (4v^2) = 2mv^2 \]
- The change in kinetic energy (\( \Delta KE \)) is:
\[ \Delta KE = KE_f - KE_i = 2mv^2 - \frac{1}{2} mv^2 = \frac{4mv^2 - mv^2}{2} = \frac{3mv^2}{2} \]

3. **Relate Work Done to Electric Field**:
- The work done by the electric field \( W \) can be expressed as:
\[ W = qE \cdot d \]
- Here, \( d \) is the distance moved in the direction of the electric field. The particle moves from (0, a) to (2a, 0), which is a distance of \( \sqrt{(2a - 0)^2 + (0 - a)^2} = \sqrt{4a^2 + a^2} = \sqrt{5a^2} = a\sqrt{5} \).

4. **Set up the equation**:
- Equating the work done to the change in kinetic energy:
\[ qE \cdot a\sqrt{5} = \frac{3mv^2}{2} \]

5. **Solve for Electric Field \( E \)**:
- Rearranging gives:
\[ E = \frac{3mv^2}{2qa\sqrt{5}} \]

### Part (b): Rate of Work Done by Both Fields at (2a, 0)

1. **Calculate the Electric Force**:
- The electric force \( \vec{F_e} \) on the particle is given by:
\[ \vec{F_e} = qE \hat{i} \]

2. **Calculate the Magnetic Force**:
- The magnetic force \( \vec{F_m} \) is given by:
\[ \vec{F_m} = q(\vec{v} \times \vec{B}) \]
- At point (2a, 0), the velocity \( \vec{v} = -2v \hat{j} \) and \( \vec{B} = B \hat{k} \):
\[ \vec{F_m} = q(-2v \hat{j} \times B \hat{k}) = -2qvB (\hat{j} \times \hat{k}) = -2qvB (-\hat{i}) = 2qvB \hat{i} \]

3. **Total Force**:
- The total force \( \vec{F} \) acting on the particle is:
\[ \vec{F} = \vec{F_e} + \vec{F_m} = qE \hat{i} + 2qvB \hat{i} = q(E + 2vB) \hat{i} \]

4. **Rate of Work Done**:
- The rate of work done \( P \) by the electric and magnetic fields is given by:
\[ P = \vec{F} \cdot \vec{v} = q(E + 2vB)(-2v \hat{j}) \cdot \hat{i} = 0 \]
- Since the electric and magnetic forces are perpendicular to the velocity, the rate of work done is zero.

### Final Answers:
- (a) Magnitude of Electric Field at point (0, a):
\[ E = \frac{3mv^2}{2qa\sqrt{5}} \]
- (b) Rate of work done by both fields at (2a, 0):
\[ P = 0 \]