a) 7 V
b) 8 V
c) -8 V
d) -7 V
Answer: c) -8 V
Solution:
\(\begin{array}{l}\vec{E}=(3\hat{i}+4y\hat{j})\end{array} \)
Ex = 3
Ey = 4y
\(\begin{array}{l}V= -\int_{0}^{2}E_{x}dx-\int_{0}^{1}E_{y}dy\end{array} \)
\(\begin{array}{l}V= -\int_{0}^{2}3dx-\int_{0}^{1}4ydy\end{array} \)
\(\begin{array}{l}= -3[x]_{0}^{2}-2[y^{2}]_{0}^{1}\end{array} \)
= – 3 x 2 – 2 x 1
= – 8 V
