Question

A room with 3.0-m-high ceilings has a metal plate on the floor with V = 0 V and a separate metal plate on the ceiling. A 1.0 g glass ball charged to +4.9 nC is shot straight up at 5.0 m/s. How high does the ball go if the ceiling voltage is (a) $+3.0\times {10}^{6}V$and (b) $-3.0\times {10}^{6}V$?

Short answer

The value of high for the ceiling voltage is $+3.0\times {10}^{6}V$is 0.85cm

The value of high for the ceiling voltage is $-3.0\times {10}^{6}V$is 2.55m

Step-by-step solution

Step: 1 The formulation of celling voltage 

The net force = $F_e+F_g=(electricforce+gravitationalforce)$

$$\begin{gathered} F_e=Eq \\ {F}_g=mg \\ {F}_e+F_g=Eq+mg \end{gathered}$$

from newton law formula

$F_{net}=ma$

applying all data

$$\begin{gathered} ma=Eq+mg \\ a=\frac{Eq+mg}{m} \\ AsE=\left(\frac{∆V}{md}\right) \\ ∆V=3\times {10}^{6}V \\ g=9.8m/s^2 \\ m=1g \\ d=3.0m \\ q=4.9nC \\ a=14.7m/s^2 \end{gathered}$$

For Trigonometric equation;

$$\begin{gathered} {v_y}^2={v_{oy}}^2+2ay \\ {v}_{oy}=0.5 \\ {v}_y=0 \\ y=0.85m \end{gathered}$$

Step: 2 The definition of celling voltage 

The net force
$F_e-F_g=(electricforce+gravitationalforce)$

As F=ma

$$\begin{gathered} ma=Eq-mg \\ a=\frac{Eq-mg}{m} \\ AsE=\left(\frac{∆V}{md}\right) \\ ∆V=3\times {10}^{6}V \\ g=9.8m/s^2 \\ m=1g \\ d=3.0m \\ q=4.9nC \\ a=-4.9m/s^2 \end{gathered}$$

From trigonometric equation;

$$\begin{gathered} {v_y}^2={v_{oy}}^2+2ay \\ {v}_{oy}=0.5 \\ {v}_y=0 \\ y=2.55m \end{gathered}$$