A tiny sphere of mass 7.70 mg and charge −2.80 nC is initially at a distance of 1.64 μm from a fixed charge of +8.69 NC.
(a) If the 7.70-mg sphere is released from rest, find its kinetic energy when it is 0.500 μm from the fixed charge.
(b) If the 7.70-mg sphere is released from rest, find its speed when it is 0.500 μm from the fixed charge.

Explanation:

We shall apply conservation of energy in an electrical field to solve the problem.

potential energy initial of sphere = 9 x 10⁹ x Q q / d , Q and q are two given charges at d distance .

Q = 8.69 x 10⁻⁹C , q = -2.8 x 10⁻⁹ C, d = 1.64 x 10⁻⁶ m

potential energy V₁ at distance 1.64 x 10⁻⁶

 = - 2.8 x 8.69 x 10⁻¹⁸x 9 x 10⁹ / 1.64 x 10⁻⁶

= - 133.53  x 10⁻³ J

potential energy V₂ at distance 0.5 x 10⁻⁶

 = - 2.8 x 8.69 x 10⁻¹⁸x 9 x 10⁹ / .5 x 10⁻⁶

= - 437.97   x 10⁻³ J

Decrease in potential energy = 304.44 x 10⁻³ J .

This will result in increase in kinetic energy

kinetic energy at .5 m = 304 x 10⁻³ J

b ) 1 / 2 m v² = kinetic energy , m is mass and v is velocity

1 / 2 m v²  = 304 x 10⁻³

v² = 2 x 304 x 10⁻³  / 7.7 x 10⁻³

= 79

v = 8.9 m /s

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

at right angles to each other