an eletrically charged point particle that has a mass of 1.50 x10^-6 kg is launc
ID: 1755196 • Letter: A
Question
an eletrically charged point particle that has a mass of 1.50 x10^-6 kg is launched horizontally into a uniform electirc fieldwith a magnitude of 925 N/C the particle's intial velocity is 8.80 m/s directed eastward and the external electic field is alsodirected eastward a short time after being launched the particle is 6.71 X 10^-3m below and 0.160m east of its initial position, determine the net charge carried by the particle including thealgebraic sign + or - the particle's intial velocity is 8.80 m/s directed eastward and the external electic field is alsodirected eastward a short time after being launched the particle is 6.71 X 10^-3m below and 0.160m east of its initial position, determine the net charge carried by the particle including thealgebraic sign + or -Explanation / Answer
the force acting on the charged point particle is F = E * q or m * a = E * q or a = (E * q/m) ----------(1) E = 925 N/C,q is the charge of the point particle and m is themass of the charged point particle and is equal to 1.50 * 10-6 kg we know that v2 - u2 = 2aS or a = (v2 - u2/2S) ----------(2) v = 0,u = 8.80 m/s and S = 0.160 m from (1) and (2) we get (v2 - u2/2S) = (E * q/m) or q = (m/E) * (v2 - u2/2S) or q = (m/E) * ((0)2 - u2/2S) = -(m *u2/2E * S) the magnitude of the charge is q = (m * u2/2E * S)and the sign of the charge is negative.Related Questions
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