Question

An electrically charged point particle with a mass of 1.15 x 10^-6 kg is launched horizontally into a uniform electric field with a magnitude of 914 N/C. The particle's initial velocity is 7.15 m/s, directed eastward, and the external electric field is also directed eastward. A short time after being launched, the particle is 8.31 x 10^-3 m below and 0.124 m east of its initial position. Determine the net charge carried by the particle, including the algebraic sign (+ or -).

Answer 1

Mass (m) = 1.15 x 10^-6 kg

Electric field (E) = 914 N/C eastward direction

Initial velocity of particle (v) = 7.15 m/s eastward direction

Position of particle after short time is 8.31 x 10^-3 m below and 0.124 m east of its initial position.

Let us suppose that the charge of the particle was q and the position was measured after a small time t.

Forces acting on particle are gravitational mg downwards and electrical qE in eastward (assuming q as +ve).

position of charge in vertical direction after time t (only vertical component contribution)

F = - mg since a = -g

s = ut + 0.5 at²   equation of motion

- 8.31 x 10^-3 = 0 x t + 0.5 x -9.81 x t²   initial velocity in vertical direction is zero and upward direction is +ve

t² = 1.694 x 10^-3

t = 0.04116 sec short time after which the measurment of position was done

position of charge in horizontal direction after time t (only horizontal component contribution)

F = qE = ma since a = qE/m

s = ut + 0.5 at²   equation of motion

0.124 = 7.15 x  0.04116 + 0.5 x [qE/m] x ( 0.04116)²

0.124 = 0.2943 + 8.471 x 10^-4 x [(q x 914) / 1.15 x 10^-6 ]

673238.73 x q = 0.124 - 0.2943

q = -0.1703 / 673238.73 = -2.5296 x 10^-7 C or -0.253 ?C

The cherge is negative with magnitude - 0.253 ?C