Question

A 2.00-g charged particle is released from rest in a region that has a uniform electric field = (160 N/C) . After traveling a distance of 0.520 m in this region, the particle has a kinetic energy 0.080 J. Determine the charge of the particle.

Answer 1

speed=v (say)
then 0.5*0.002*v^2=0.8
v=28.284 m/s
acceleration=final speed^2/(2*distance travelled)=769.23 m/s^2
force=mass*acceleration=1.5384 N
now electrostatic force=charge*field
so if charge is q,
then q*160=1.5384
q=9.6 mC

Answer 2

mass of the charged particle is m = 2.00 g = 2.00 * 10^-3 kg
the electric field is E = 160 N/C
the distance travelled by charge is d = 0.520 m
the kinetic energy is K = 0.080 J
we know that
K = (1/2)mv^2
where v is speed of charge
or v^2 = (2K/m)
or v = (2K/m)^1/2
let the acceleration of charge be a
we know that
v^2 - u^2 = 2ad
where u = 0
or v^2 = 2ad
or a = v^2/2d
the force acting on the charge is
F = m * a
the above force is equal to the electric force acting on the charge therefore
F_e = E * q
where q is charge of charged particle
Here,F = F_e
or E * q = m * a
or q = (m * a/E)

Answer 3

speed=v (say)
then 0.5*0.002*v^2=0.8
v=28.284 m/s
acceleration=final speed^2/(2*distance travelled)=769.23 m/s^2
force=mass*acceleration=1.5384 N
now electrostatic force=charge*field
so if charge is q,
then q*160=1.5384
q=9.6 mC