An electrically charged point particle with a mass of 1.41 x 10-6 kg is launched horizontally into a uniform electric field with a magnitude of 686 N/C. The particle's initial velocity is 5.39 m/s, directed eastward, and the external electric field is also directed eastward. A short time after being launched, the particle is 9.89 x 10-3 m below and 0.151 m east of its initial position. Determine the net charge carried by the particle, including the algebraic sign ( or -).

the net charge carried by the particle is -2.35 × 10⁻⁷ C

Explanation:

Given that;

mass m = 1.41 × 10⁻⁶ kg

Magnitude E = 686 N/C

v = 5.93 m/s

A short time after being launched, the particle is 9.89 x 10-3 m below and 0.151 m east of its initial position

consider vertical;

u =0, y = 9.89 × 10⁻³ m, g = a = 9.8 m/s²

Now, from the second equation of motion;

s = ut + 1/2 × at²

so, y = ut + 1/2 × at²

we substitute

9.89 × 10⁻³ = 0×t + (1/2 × 9.8 × t²)

9.89 × 10⁻³ = 4.9t²

t² = 9.89 × 10⁻³ / 4.9

t² = 0.0201836

t = √0.00201836

t = 0.0449 s

Consider horizontal;

u = 5.93 m/s, a = (qE/m) m/s, t = 0.0449 s, x = 0.151 m

also, from the second equation of motion;

s = ut + 1/2 × at²

so

x =  ut + 1/2 ×  (qE/m) × t²

we substitute

0.151 =  (5.93 × 0.0449) + (1/2 ×  ((q×686)/1.41 × 10⁻⁶) × (0.0449 )²

0.151 = 0.266257 +  0.69149143q / 1.41 × 10⁻⁶

0.151 - 0.266257 =  0.69149143q / 1.41 × 10⁻⁶

-0.115257 = 0.69149143q / 1.41 × 10⁻⁶

-0.115257 × 1.41 × 10⁻⁶ = 0.69149143q

0.69149143q = -1.6251237 × 10⁻⁷

q = -1.6251237 × 10⁻⁷ / 0.69149143

q = -2.35 × 10⁻⁷ C

Therefore, the net charge carried by the particle is -2.35 × 10⁻⁷ C

probably not if net force is zero

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

                                              ⇒      

here   k is the proportionality constant whose value is 1.

                                                            [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.