Not really sure how do this problem or even where tobegin... An electrically cha
ID: 1671516 • Letter: N
Question
Not really sure how do this problem or even where tobegin...An electrically charged point particle that has a mass of 1.5x 10^-6 kg is launched into a uniform electric field with amagnitude of 925 N/C. The particle'e initial velocity is 8.80m/s,directed eastward, and the external electric field is also directedeastward. A short time after ebing launched, the particle is 6.71 x10^-3 m below and 0.160m east of its initial position. Determinethe net charge carried by the particle, including the algebraicsign (+ or -). Not really sure how do this problem or even where tobegin...
An electrically charged point particle that has a mass of 1.5x 10^-6 kg is launched into a uniform electric field with amagnitude of 925 N/C. The particle'e initial velocity is 8.80m/s,directed eastward, and the external electric field is also directedeastward. A short time after ebing launched, the particle is 6.71 x10^-3 m below and 0.160m east of its initial position. Determinethe net charge carried by the particle, including the algebraicsign (+ or -).
Explanation / Answer
An electrically charged point particle that has a mass of 1.5x 10^-6 kg is launched into a uniform electric field with amagnitude of 925 N/C. The particle's initial velocity is 8.80m/s,directed eastward, and the external electric field is also directedeastward. A short time after being launched, the particle is 6.71 x10^-3 m below and 0.160m east of its initial position. Determinethe net charge carried by the particle, including the algebraicsign (+ or -).the time taken by the particle to reach the point below=t
using the equation
h=0+1/2g*t*t
6.71*10-3=0+4.9*t*t
t=3.70*10-2sec
in the same time the particle reaches a horizontal distance of0.160m
let the electric charge present on the particle be q
the force due to electric field=Eq=F=925q
so 0.160=8.8*3.70*10-2+1/2 (F/mass)(3.70*10-2)2=
0.160 =8.8*3.70*10-2+1/2*925q /1.5 x10-6 *(3.70*10-2)2
calculating we will get the magnitude of the charge.
as the charge is moved in the direction of the field the charge ispositive.