Question

A 2.0 g ball with a +0.05 C charge is moved from the negative plate to the positive plate, then released. The potential difference between the plates is 12.0V. When the ball strikes the (–) plate its velocity is most nearly equal to:__________
a. 6 m/s
b. 12 m/s
c. 17 m/s
d. 25 m/s

Answer 1

Answer:

25 m/s

Explanation:

The ball will have kinetic energy that is proportional to the energy due to the product of the charge and the potential difference.

mass of the ball m = 2.0 g = 2 x $10^{-3}$ kg

charge on the ball C = +0.05 C

potential difference on plates V = 12.0 V

The energy due to the electric potential difference and the charge will be

E = CV

E = 0.05 X 12 = 0.6 Joules

The ball will move with a kinetic energy that is proportional to this energy calculated (due to energy conservation principle), therefore,

E = $\frac{1}{2}$$mv^{2}$ ..........which is the kinetic energy relationship where v is the velocity of the ball, substituting values in, we have

0.6 = $\frac{1}{2}$ x 2 x $10^{-3}$ x $v^{2}$

0.6 = $10^{-3}$ x $v^{2}$

600 = $v^{2}$

v = $\sqrt{600}$ = 24.49 ≅ 25 m/s