(a) To find the strength of the electric field, we first need to find the acceleration of the electron. We know that the electron comes to a stop, so its final velocity is 0 m/s. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Rearranging for a gives a = (v^2 - u^2) / (2s). Substituting the given values gives a = (0 - (5.00 × 10^6)^2) / (2 * 0.20) = -6.25 × 10^15 m/s^2. The acceleration is negative because the electron is slowing down.

The force on the electron due to the electric field is F = ma, where m is the mass of the electron (9.11 × 10^-31 kg). Substituting the values gives F = (9.11 × 10^-31 kg) * (-6.25 × 10^15 m/s^2) = -5.69 × 10^-15 N.

The strength of the electric field is E = F/q, where q is the charge of the electron (1.60 × 10^-19 C). Substituting the values gives E = (-5.69 × 10^-15 N) / (1.60 × 10^-19 C) = 3.56 × 10^4 N/C.

(b) The electron stops momentarily because it has lost all its kinetic energy due to the work done by the electric field. The electric field exerts a force on the electron in the opposite direction to its motion, causing it to decelerate. When the electron's kinetic energy has been completely converted into potential energy, it momentarily comes to a stop before being accelerated back in the opposite direction by the electric field.

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(a) To find the strength of the electric field, we first need to find the acceleration of the electron. We know that the electron comes to a stop, so its final velocity is 0 m/s. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Rearranging for a gives a = (v^2 - u^2) / (2s). Substituting the given values gives a = (0 - (5.00 × 10^6)^2) / (2 * 0.20) = -6.25 × 10^15 m/s^2. The acceleration is negative because the electron is slowing down.

The force on the electron due to the electric field is F = ma, where m is the mass of the electron (9.11 × 10^-31 kg). Substituting the values gives F = (9.11 × 10^-31 kg) * (-6.25 × 10^15 m/s^2) = -5.69 × 10^-15 N.

The strength of the electric field is E = F/q, where q is the charge of the electron (1.60 × 10^-19 C). Substituting the values gives E = (-5.69 × 10^-15 N) / (1.60 × 10^-19 C) = 3.56 × 10^4 N/C.

(b) The electron stops momentarily because it has lost all its kinetic energy due to the work done by the electric field. The electric field exerts a force on the electron in the opposite direction to its motion, causing it to decelerate. When the electron's kinetic energy has been completely converted into potential energy, it momentarily comes to a stop before being accelerated back in the opposite direction by the electric field.

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(a) To find the strength of the electric field, we first need to find the acceleration of the electron. We know that the electron comes to a stop, so its final velocity is 0 m/s. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Rearranging for a gives a = (v^2 - u^2) / (2s). Substituting the given values gives a = (0 - (5.00 × 10^6)^2) / (2 * 0.20) = -6.25 × 10^15 m/s^2. The acceleration is negative because the electron is slowing down.

The force on the electron due to the electric field is F = ma, where m is the mass of the electron (9.11 × 10^-31 kg). Substituting the values gives F = (9.11 × 10^-31 kg) * (-6.25 × 10^15 m/s^2) = -5.69 × 10^-15 N.

The strength of the electric field is E = F/q, where q is the charge of the electron (1.60 × 10^-19 C). Substituting the values gives E = (-5.69 × 10^-15 N) / (1.60 × 10^-19 C) = 3.56 × 10^4 N/C.

(b) The electron stops momentarily because it has lost all its kinetic energy due to the work done by the electric field. The electric field exerts a force on the electron in the opposite direction to its motion, causing it to decelerate. When the electron's kinetic energy has been completely converted into potential energy, it momentarily comes to a stop before being accelerated back in the opposite direction by the electric field.

An electron moves in the direction of a uniform electric field. Its initial speed is5.00 × 106 m s−1 and stops momentarily after moving through a distance of 20.0 cm.(a) What is the strength of the electric field?(b) Why the electron stops momentarily?

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