An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distance of 1.2 cm. What is the electric field strength?

The work done (W) on a charged particle moving through an electric field is given by the equation:

\[ W = \Delta KE \]

where \( W \) is the work done, and \( \Delta KE \) is the change in kinetic energy of the charged particle.

The change in kinetic energy (\( \Delta KE \)) can be calculated using the equation:

\[ \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \]

where \( m \) is the mass of the charged particle, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.

Since the electron is moving parallel to the electric field, the work done by the electric field is equal to the change in kinetic energy. The work done by the electric field (\( W \)) can also be expressed as:

\[ W = q \cdot E \cdot d \]

where \( q \) is the charge of the electron, \( E \) is the electric field strength, and \( d \) is the distance over which the electron moved.

Setting the two expressions for work equal to each other, we get:

\[ q \cdot E \cdot d = \frac{1}{2} m (v_f^2 - v_i^2) \]

We can solve for \( E \) (electric field strength):

\[ E = \frac{1}{q} \cdot \frac{1}{2m} \cdot (v_f^2 - v_i^2) \cdot \frac{1}{d} \]

Given:

- \( q \) (charge of the electron) is the elementary charge \( e \), approximately \( 1.602 \times 10^{-19} \) C.

- \( m \) (mass of the electron) is approximately \( 9.109 \times 10^{-31} \) kg.

- \( v_i \) (initial velocity) is \( 2.0 \times 10^7 \) m/s.

- \( v_f \) (final velocity) is \( 4.0 \times 10^7 \) m/s.

- \( d \) (distance) is \( 1.2 \) cm, which needs to be converted to meters (\( 0.012 \) m).

Plug in the values and calculate \( E \):

\[ E = \frac{1}{1.602 \times 10^{-19}} \cdot \frac{1}{2 \times 9.109 \times 10^{-31}} \cdot ((4.0 \times 10^7)^2 - (2.0 \times 10^7)^2) \cdot \frac{1}{0.012} \]

This calculation will give you the electric field strength in N/C (Newtons per Coulomb).