Question

A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t = 0.0420 s a constant electric field of magnitude E = 7.80 × 102 N/C is directed vertical to the ball which makes the ball rise to a height of d. After this time the electric field is turned off and the ball returns to the surface. Find the height in meters which the ball is able to be lifted off the surface.

Answer 1

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

Electric field $E=7.80\times10^{2}\ N/C$

We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

We need to calculate the acceleration

Using newton's law

$F = ma$

$a = \dfrac{F}{m}$

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m