180- In a uniform electric field $E = 10^5\ \frac{\text{N}}{\text{C}}$, a particle with electric charge $q = -5\ \mu\text{C}$ moves along path $ABC$ from $A$ to $C$. How does the electric potential energy of the particle change along this path?
$$\left(\sin\alpha = 0.8\ ,\ AB = BC = 5\ \text{cm}\right)$$
[Figure: Uniform electric field $\vec{E}$ pointing to the right, with path ABC where A is on the left, B is in the middle, and C is at the bottom right, with angle $\alpha$ at B.]\begin{flushright} (1) $0.1$ joule, increases
(2) $0.1$ joule, decreases
(3) $0.4$ joule, increases
(4) $0.4$ joule, decreases \end{flushright}
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