P19B2015 – Plastic Plate & Charged Ball

The large plastic plate creates an electric field which causes the plastic ball to experience the electric force $F_E$ and therefore an acceleration.

The electric field created by the large plastic plate is perpendicular to the plate and uniform. It is equal everywhere to $E=\sigma /2{\varepsilon }_0$ and points away from the plate (perpendicularly).

The small plastic ball therefore experiences the electric force $F_E=qE$ in addition to its weight force, and Newton’s Second Law yields

\begin{aligned}
\left\{ \begin{array}{c}
F_{net\ x}=ma_x \\
F_{net\ y}=ma_y \end{array}
\right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
qE=ma_x \\
-mg=ma_y \ \ \ \ \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{\frac{q\sigma }{2{\varepsilon }_0}=ma_x} \\
-g=a_y \ \ \ \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{a_x=\frac{qQ}{2Am{\varepsilon }_0}} \\
a_y=-g \ \ \ \ \ \ \  \end{array}
\right.
\end{aligned}

Therefore the magnitude of the acceleration of the ball is given by

\boxed{a=\sqrt{a^2_x+a^2_y}=\sqrt{{\left(\frac{qQ}{2Am{\varepsilon }_0}\right)}^2+g^2}=\sqrt{{\left(\frac{0.3\cdot 4}{2\cdot {\varepsilon }_0\cdot 0.5\cdot 1}\right)}^2+{9.8}^2}=\sqrt{{\left(\frac{1.2}{{\varepsilon }_0}\right)}^2+{9.8}^2}\ \ \ m/s^2}

The acceleration of the ball is

\boxed{a=\sqrt{{\left(\frac{1.2}{{\varepsilon }_0}\right)}^2+{9.8}^2}\approx \frac{1.2}{{\varepsilon }_0}}

Note: this value is ridiculously large because the values of charge chosen are ridiculously large ($1\ C$ is HUGE).

Note: given the extremely small value of ${\varepsilon }_0$, the fraction ${\left(1.2/{\varepsilon }_0\right)}^2$ will dwarf ${9.8}^2$ and you can write that

\boxed{a\approx \frac{1.2}{{\varepsilon }_0}\ \ \ m/s^2}

This comes from the fact that the values of charge used are huge which causes the electric force to be orders of magnitude bigger than the weight force. This allows you to reasonably neglect the weight force in front of the electric force.

Answer: b

Newton’s Third Law ensures that the force of the plate on the ball is the same in magnitude as the force of the ball on the plate.