Answer:
13.4 x 10 raise to power -19 C
Explanation:
. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m
. The uniform electric field is E = 214 N/M
, The decrease in electrical potential energy is
d(P.E) = 51.63 x 10 raise to power -19 J
Let the magnitude of the charge of the moving particle be q
which is given by the equation
d(P.E) =qEd
51.63 x 10 power -19 = q(214)(0.018)
51.63 x 10 power -19 =3.852q
by making q the formular,
q = 13.4 x 10 power -19 C
how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?
a. it is reduced by a factor of 3
Explanation:
s = ut + 1/2 a t^2
200 = 0 * 6 + 1/2 * a * (6)^2
200 = 1/2 * a * 36
200 = 18 a
a = 200/18
a= 11.1m/sec^2
v = u + at
v = 0 + 11.1 * 6
v = 66.6m/s
hope it helps you
The answer is either c or d but c is the best answer