The problem involves a charged particle moving in an electric field. The force on the charged particle due to the electric field is given by F = qE, where q is the charge and E is the electric field strength. This force will cause the particle to accelerate, changing its velocity.

The change in velocity can be broken down into two components: one parallel to the initial velocity (which doesn't change the direction of the velocity, only its magnitude), and one perpendicular to the initial velocity (which changes the direction of the velocity, but not its magnitude). The latter is responsible for the 60 degree deviation in the particle's path.

The component of the acceleration perpendicular to the initial velocity is given by a_perp = F/m = qE/m. This causes the particle to move in a circular path, with the radius of the circle given by r = v0^2/a_perp = mv0/qE.

The angle by which the particle deviates is given by theta = v_final/v0 = 2, where v_final is the final velocity of the particle. This can be related to the time for which the field acts and the radius of the circular path by the equation theta = v0*t/r.

Substituting the expressions for theta and r into this equation gives 2 = v0*t/(mv0/qE), which simplifies to 2 = t*qE/mv0. Solving for t gives t = 2mv0/qE.

The problem asks for the value of x^2, where x = t/(mv0/qE) = 2. Therefore, x^2 = 4.

Question

The problem involves a charged particle moving in an electric field. The force on the charged particle due to the electric field is given by F = qE, where q is the charge and E is the electric field strength. This force will cause the particle to accelerate, changing its velocity.

The change in velocity can be broken down into two components: one parallel to the initial velocity (which doesn't change the direction of the velocity, only its magnitude), and one perpendicular to the initial velocity (which changes the direction of the velocity, but not its magnitude). The latter is responsible for the 60 degree deviation in the particle's path.

The component of the acceleration perpendicular to the initial velocity is given by a_perp = F/m = qE/m. This causes the particle to move in a circular path, with the radius of the circle given by r = v0^2/a_perp = mv0/qE.

The angle by which the particle deviates is given by theta = v_final/v0 = 2, where v_final is the final velocity of the particle. This can be related to the time for which the field acts and the radius of the circular path by the equation theta = v0*t/r.

Substituting the expressions for theta and r into this equation gives 2 = v0*t/(mv0/qE), which simplifies to 2 = t*qE/mv0. Solving for t gives t = 2mv0/qE.

The problem asks for the value of x^2, where x = t/(mv0/qE) = 2. Therefore, x^2 = 4.

Knowee AI · Accepted Answer

The problem involves a charged particle moving in an electric field. The force on the charged particle due to the electric field is given by F = qE, where q is the charge and E is the electric field strength. This force will cause the particle to accelerate, changing its velocity.

The change in velocity can be broken down into two components: one parallel to the initial velocity (which doesn't change the direction of the velocity, only its magnitude), and one perpendicular to the initial velocity (which changes the direction of the velocity, but not its magnitude). The latter is responsible for the 60 degree deviation in the particle's path.

The component of the acceleration perpendicular to the initial velocity is given by a_perp = F/m = qE/m. This causes the particle to move in a circular path, with the radius of the circle given by r = v0^2/a_perp = mv0/qE.

The angle by which the particle deviates is given by theta = v_final/v0 = 2, where v_final is the final velocity of the particle. This can be related to the time for which the field acts and the radius of the circular path by the equation theta = v0*t/r.

Substituting the expressions for theta and r into this equation gives 2 = v0*t/(mv0/qE), which simplifies to 2 = t*qE/mv0. Solving for t gives t = 2mv0/qE.

The problem asks for the value of x^2, where x = t/(mv0/qE) = 2. Therefore, x^2 = 4.

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