The electric potential V(x) in a region aroun | vedclass

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(C) The electric field $E$ is given by the negative gradient of the potential: $E = -\frac{dV}{dx} = -\frac{d}{dx}(4x^2) = -8x 	ext{ V/m}$.According

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(C) The electric field $E$ is given by the negative gradient of the potential: $E = -\frac{dV}{dx} = -\frac{d}{dx}(4x^2) = -8x 	ext{ V/m}$.According to Gauss's Law,the total flux $\phi$ through a closed surface is $\phi = \oint E \cdot dA = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.For a cube of side $L = 1 	ext{ m}$ centered at the origin,the faces are at $x = 0.5 	ext{ m}$ and $x = -0.5 	ext{ m}$.The electric field at $x = 0.5 	ext{ m}$ is $E_1 = -8(0.5) = -4 	ext{ V/m}$ (pointing towards the origin).The electric field at $x = -0.5 	ext{ m}$ is $E_2 = -8(-0.5) = 4 	ext{ V/m}$ (pointing away from the origin).The flux through the face at $x = 0.5 	ext{ m}$ is $\phi_1 = E_1 \cdot A = -4 	imes (1^2) = -4 	ext{ V} \cdot 	ext{m}^2$.The flux through the face at $x = -0.5 	ext{ m}$ is $\phi_2 = E_2 \cdot A = 4 	imes (1^2) = 4 	ext{ V} \cdot 	ext{m}^2$.The flux through the other four faces is zero because the electric field is parallel to these faces.The net flux is $\phi_{	ext{net}} = \phi_1 + \phi_2 = -4 + 4 = 0$.Since $\phi_{	ext{net}} = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$,we have $q_{	ext{enclosed}} = 0$.

The electric potential $V(x)$ in a region around the origin is given by $V(x) = 4x^2 \text{ V}$. The electric charge enclosed in a cube of $1 \text{ m}$ side with its centre at the origin is (in coulomb):

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