Question:

Given \( E(t) = 108t - 22 \), how is velocity derived from the electric field?

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To find the velocity from the electric field, integrate the electric field with respect to time, considering the charge and mass of the particle.
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Approach Solution - 1

Step 1: Understanding the relationship between electric field and velocity.
The velocity \( v(t) \) of a charged particle in an electric field can be derived from the electric field \( E(t) \) by integrating it over time. The relationship is given by: \[ v(t) = \frac{q}{m} \int E(t) \, dt \] where \( q \) is the charge of the particle, \( m \) is the mass, and \( E(t) \) is the electric field as a function of time.
Step 2: Substituting the given electric field.
We are given \( E(t) = 108t - 22 \), so we substitute this into the equation: \[ v(t) = \frac{q}{m} \int (108t - 22) \, dt \] Step 3: Performing the integration.
Integrating \( 108t - 22 \) with respect to time: \[ \int (108t - 22) \, dt = \frac{108t^2}{2} - 22t + C \] Step 4: Final expression for velocity.
Thus, the velocity as a function of time is: \[ v(t) = \frac{q}{m} \left( \frac{108t^2}{2} - 22t + C \right) \] Step 5: Conclusion.
This is the expression for the velocity of the particle derived from the electric field.
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Approach Solution -2

Step 1: A charged particle's velocity from an electric field is found by integrating: v(t) = (q/m) × ∫E(t) dt.

Step 2: Here E(t) = 108t − 22.

Step 3: Integrating gives ∫(108t − 22) dt = 54t² − 22t + C.

Step 4: So the velocity is v(t) = (q/m) × (54t² − 22t + C).
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Approach Solution -3

Two-stage approach: field to acceleration, then acceleration to velocity.
By Newton's second law, the electric force on the charge is \( F = qE(t) \), so its acceleration is \[ a(t) = \frac{qE(t)}{m} = \frac{q}{m}(108t - 22) \]
Velocity is the time-integral of acceleration, taken from an initial time to time \( t \), starting from some initial velocity \( v_0 \): \[ v(t) = v_0 + \int_0^t a(t')\,dt' = v_0 + \frac{q}{m}\left(54t^2 - 22t\right) \] which is the velocity at time \( t \), written with an explicit initial condition rather than a free constant.
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