An electron moving to the right at 7.5 x 10⁵ m/s enters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 5.0 cm,
(a) what direction is required for the electric field, and
(b) what is the strength of the field?

검증된 단계별 안내

Determine the direction of the electric field: Since the electron is negatively charged, the electric field must point in the opposite direction of its motion to exert a force that decelerates it. This is because the force on a charged particle in an electric field is given by \( \mathbf{F} = q \mathbf{E} \), where \( q \) is the charge of the particle and \( \mathbf{E} \) is the electric field vector.

Use kinematic equations to find the required deceleration: The electron starts with an initial velocity \( v_0 = 7.5 \times 10^5 \ \text{m/s} \), comes to rest (\( v = 0 \)), and travels a distance \( d = 5.0 \ \text{cm} = 0.050 \ \text{m} \). Use the kinematic equation \( v^2 = v_0^2 + 2a d \) to solve for the acceleration \( a \). Rearrange to get \( a = -\frac{v_0^2}{2d} \).

Relate the acceleration to the electric field: The force on the electron is \( F = ma \), where \( m \) is the mass of the electron. The force is also given by \( F = qE \), where \( q \) is the charge of the electron. Equating these, \( ma = qE \), solve for \( E \) to get \( E = \frac{ma}{q} \).

Substitute known values into the equations: Use the mass of the electron \( m = 9.11 \times 10^{-31} \ \text{kg} \), the charge of the electron \( q = 1.60 \times 10^{-19} \ \text{C} \), and the acceleration \( a \) calculated in step 2 to find the magnitude of the electric field \( E \).

Combine the results: The direction of the electric field is opposite to the electron's motion, and the magnitude of the electric field is determined from the calculations in step 4. Ensure all units are consistent and verify the results.

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주요 개념

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Electric Field Direction

The direction of an electric field is defined as the direction a positive test charge would move. In this scenario, since the electron is moving to the right and needs to be brought to rest, the electric field must be directed to the left. This opposing direction is necessary to exert a force on the negatively charged electron, decelerating it until it stops.

Force on a Charged Particle

The force experienced by a charged particle in an electric field is given by F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength. For an electron, which has a negative charge, the force will act in the opposite direction to the electric field. This relationship is crucial for determining how the electric field will affect the electron's motion.

Kinematics and Deceleration

Kinematics involves the equations of motion that describe how an object moves under various forces. In this case, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance to find the required electric field strength. The electron's initial velocity, the distance it must travel to stop, and the need for deceleration are key factors in calculating the necessary electric field strength.