P20B2023 – Charge Hanging from a String
Charge Hanging from a String
A very large vertical plane is uniformly covered with negative electric charge whose distribution per unit area is $\sigma < 0$. A point mass of mass $m$ and charge $q < 0$ is attached to a string hanging down from the ceiling.
1. Determine the direction and magnitude of the electric field created by the vertical plane.
View answerHide answerTreating the very large vertical plane as an infinite sheet with surface charge density $\sigma $, we know by Gauss’s Law that the electric field will point toward the plane (because $\sigma < 0$) and be equal to
\boxed{{\overrightarrow{E}}_p=\frac{\sigma }{2{\varepsilon }_0}\ \ \hat{x}}Note: because $\sigma < 0$, the electric field ${\overrightarrow{E}}_p$ does, in fact, point to the left, toward the sheet.
2. Determine the magnitude and direction of the electric force acting on the point charge.
View answerHide answerThe negative point charge will be repelled by the negative charged plane and the electric force ${\overrightarrow{F}}_E=q{\overrightarrow{E}}_p$ will point to the right as shown below.
The magnitude of the electric force acting on the negative point charge is equal to
\boxed{F_E=\left|qE_P\right|=\left|\frac{q\sigma }{2{\varepsilon }_0}\right|=\frac{q\sigma }{2{\varepsilon }_0}}where $q\sigma > 0$ because $q < 0$ and $\sigma < 0$.
3. Draw the free-body diagram of the point charge once it has reached static equilibrium.
View answerHide answerThe free-body diagram of the point charge once it has reached its equilibrium position is drawn below
4. Determine the magnitude and direction (angle from the vertical direction) of the tension force generated in the string.
View answerHide answerWe apply the Equilibrium Condition to the point charge and derive the following equations
\begin{aligned}
\left\{ \begin{array}{c}
F_{net\ x}=0 \\
F_{net\ y}=0 \end{array}
\right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{T{\mathrm{sin} \left(\theta \right)\ }-\frac{q\sigma }{2{\varepsilon }_0}=0} \\
\\
T{\mathrm{cos} \left(\theta \right)\ }-mg=0 \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{T{\mathrm{sin} \left(\theta \right)\ }=\frac{q\sigma }{2{\varepsilon }_0}} \\
\\
T{\mathrm{cos} \left(\theta \right)\ }=mg \end{array}
\right.
\end{aligned}To derive the magnitude of the tension force, we note that $T_x=T{\mathrm{cos} \left(\theta \right)\ }$ and $T_y=T{\mathrm{sin} \left(\theta \right)\ }$ and therefore write
\boxed{T=\sqrt{T^2_x+T^2_y}=\sqrt{{\left(T{\mathrm{cos} \left(\theta \right)\ }\right)}^2+{\left(T{\mathrm{sin} \left(\theta \right)\ }\right)}^2}=\sqrt{{\left(mg\right)}^2+{\left(\frac{q\sigma }{2{\varepsilon }_0}\right)}^2}}To derive the direction of the tension force, we divide the second equation by the first as follows
\begin{aligned}
\frac{T{\mathrm{sin} \left(\theta \right)\ }}{T{\mathrm{cos} \left(\theta \right)\ }}=\frac{q\sigma }{2{\varepsilon }_0mg}\ \ \ \ &\Rightarrow \ \ \ \ \ {\mathrm{tan} \left(\theta \right)\ }=\frac{q\sigma }{2{\varepsilon }_0mg} \\
\\
&\Rightarrow \ \ \ \ \ \boxed{\theta ={{\mathrm{tan}}^{-1} \left(\frac{q\sigma }{2{\varepsilon }_0mg}\right)\ }}
\end{aligned}where $\theta $ is measured counterclockwise from the vertical.