77. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); There are two contributions to the surface integral: one at the side of the rectangle at \(\displaystyle x=0\) and the other at the side at \(\displaystyle...77. \(\displaystyle Φ=\frac{q_{enc}}{ε_0}\); There are two contributions to the surface integral: one at the side of the rectangle at \(\displaystyle x=0\) and the other at the side at \(\displaystyle x=2.0m\); \(\displaystyle −E(0)[1.5m^2]+E(2.0m)[1.5m^2]=\frac{q_{enc}}{ε_0}=−100Nm2^/C\)
The diagonal is \(\displaystyle \sqrt{2}a\) and the components of the force due to the diagonal charge has a factor \(\displaystyle cosθ=\frac{1}{\sqrt{2}}\); \(\displaystyle \vec{F}_{net}=[k\frac{q^2...The diagonal is \(\displaystyle \sqrt{2}a\) and the components of the force due to the diagonal charge has a factor \(\displaystyle cosθ=\frac{1}{\sqrt{2}}\); \(\displaystyle \vec{F}_{net}=[k\frac{q^2}{a^2}+k\frac{q^2}{2a^2}\frac{1}{\sqrt{2}}]\hat{i}−[k\frac{q^2}{a^2}+k\frac{q^2}{2a^2}\frac{1}{\sqrt{2}}]\hat{j}\) If the \(\displaystyle q_2\) is to the right of \(\displaystyle q_1\), the electric field vector from both charges point to the right.
