The electric flux is given by Gauss's Law:
\(\Phi = \oint \vec{E} \cdot d\vec{A}.\)
The field $\vec{E} = 2x \hat{i}$ varies with $x$.
For the cube, only the left ($x = 0$) and right ($x = 2$) faces contribute: \(\Phi = E_\text{right} A - E_\text{left} A.\)
Substituting $A = 4 \, \mathrm{m}^2$, $E_\text{right} = 2(2) = 4$, and $E_\text{left} = 2(0) = 0$:
\(\Phi = 4 \cdot 4 - 0 \cdot 4 = 16 \, \mathrm{Nm}^2/\mathrm{C}.\)
LIST I | LIST II | ||
A | Gauss's Law in Electrostatics | I | \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\) |
B | Faraday's Law | II | \(\oint \vec{B} \cdot d \vec{A}=0\) |
C | Gauss's Law in Magnetism | III | \(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\) |
D | Ampere-Maxwell Law | IV | \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) |
The net current flowing in the given circuit is ___ A.
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .