Energy Problems

3 Point Charges

$q=5.4\mu C$

$a=5.4m$

Enter an expression for the electric potential at the origin, $V_0$, using the given symbols.

$$V=\frac{kQ}{r}$$ $$V_0=k(\frac{Q}{a}+\frac{Q}{a}+\frac{2Q}{a})$$ $$V_0=\frac{4kQ}{a}$$

Solve for the numerical value of $V_0$ in volts.

$$V_0=\frac{4(9\cdot 10^9)(5.4\cdot 10^{-6})}{5.4}$$ $$V_0=\frac{4\cdot 5.4\cdot 9\cdot 10^3}{5.4}=36000V$$

Using $a$ to denote the direction of the electric field at the origin $E_0$, measured counterclockwise from the positive x-axis, find $\tan (a)$.

$$\vec{E}=k_e\frac{Q}{r^2}\hat{r}$$ $$\vec{E}=\frac{k_e}{r^2}(Q_1\widehat{r_1}+Q_2\widehat{r_2}+Q_3\widehat{r_3})$$ $$\vec{E}=\frac{k_e}{r^2}\left(Q\left[\begin{array}{c}-a\\ 0\end{array}\right]+2Q\left[\begin{array}{c}a\\ 0\end{array}\right]+Q\left[\begin{array}{c}0\\ a\end{array}\right]\right)$$ $$\vec{E}=\frac{k_e}{r^2}Q\left[\begin{array}{c}a\\ a\end{array}\right]$$ $$\vec{E}=\left[\begin{array}{c}\frac{k_{e}Q}{a}\\ \frac{k_{e}Q}{a}\end{array}\right]$$ $$\frac{k_{e}Q}{a}=\frac{9\cdot 10^9\cdot 5.4\cdot 10^{-6}}{5.4}=9000$$ $$\vec{E}=\left[\begin{array}{c}9000\\ 9000\end{array}\right]$$ $$\tan (a)=1$$

Find $\sin (a)$.

$$\sin (a⚡)=\frac{9000}{|\vec{E}|}=\frac{9000}{\sqrt{9000^2\cdot 2}}=\frac{\sqrt{2}}{2}\approx 0.707$$

This was supposed to be $-\frac{\sqrt{2}}{2}$

Find the magnitude of $E_0$, at the origin, in newtons per coulomb.

$$-\sqrt{2}\frac{k_{e}Q}{a^2}$$ $$-\sqrt{2}\frac{9\cdot 10^9\cdot 5.4\cdot 10^{-6}}{5.4^2}$$

Two charges on a line

If you assume the potential vanishes at infinity, the potential will be zero at some point $(x_0,0)$ with $x_1<x_0<x_2$. Write an expression for $x_0$ in terms of $q_1,q_{2}x1$ and $x2$.

$$V=k_e\left(\frac{q_1}{x_1-x}+\frac{q_2}{x_2-x}\right)$$ $$0=\frac{q_1}{|x_1-x|}+\frac{q_2}{|x_2-x|}$$ $$x=\frac{q_1{x}_2+q_2{x}_1}{q_1+q_2}$$

Moving charges

How much work is done moving a proton from (5,5) to (0,5) when a charge of 4.6uC is at (0,0)

R went from $\sqrt{50}$ to 5.

$$W=-q(V_f-V_{{}_i})$$ $$V=\frac{kQ}{r}$$ $$V_i=\frac{9\cdot 4.6\cdot 10^3}{\sqrt{50}}=5854.8$$ $$V_f=\frac{9\cdot 4.6\cdot 10^3}{5}=8280$$ $$W=-(1.602\cdot 10^{-19})(2425.15)=-3.885\cdot 10^{-16}$$