Question

Consider a uniform electric field $\vec E = 3\times10^{3}\,\hat\imath\,\text{N/C}$.
(a) What is the flux of this field through a square of $10\,\text{cm}$ on a side whose plane is parallel to the $yz$-plane?
(b) What is the flux through the same square if the normal to its plane makes a $60°$ angle with the $x$-axis?

Hint:

Use flux = E A cos(theta). Plane parallel to yz means normal along x, so theta = 0 and flux = EA. For theta = 60 degrees multiply by cos 60 = 0.5. Area = (0.10 m)^2 = 1e-2 m^2.

Answer 1

The uniform field is \(\vec E = 3\times10^{3}\,\hat\imath\,\text{N/C}\). The square has side \(L = 10\,\text{cm} = 0.10\,\text{m}\), so its area is

\[A = L^{2} = (0.10)^{2} = 1\times10^{-2}\,\text{m}^{2}.\]

Step 1: Concept of flux. The electric flux through a flat area is

\[\phi = \vec E \cdot \vec A = E\,A\,\cos\theta,\]

where \(\theta\) is the angle between \(\vec E\) and the outward normal \(\vec A\) to the surface.

Step 2: Part (a) plane parallel to the yz-plane. If the plane of the square is parallel to the yz-plane, its normal points along the x-axis, i.e. along \(\hat\imath\), so \(\theta = 0^\circ\) and \(\cos\theta = 1\).

\[\phi_a = E\,A = (3\times10^{3})(1\times10^{-2})\]\[\phi_a = 30\,\text{N m}^{2}/\text{C}\]

Step 3: Part (b) normal at \(60^\circ\) to the x-axis. Now \(\theta = 60^\circ\), so \(\cos 60^\circ = 0.5\).

\[\phi_b = E\,A\,\cos 60^\circ = (3\times10^{3})(1\times10^{-2})(0.5)\]\[\phi_b = 15\,\text{N m}^{2}/\text{C}\]\[\boxed{\phi_a = 30\,\text{N m}^{2}/\text{C}, \quad \phi_b = 15\,\text{N m}^{2}/\text{C}}\]

Answer 2

Projected-area viewpoint:

Step 1: Flux equals the field magnitude times the area projected perpendicular to the field. The field lies entirely along \(\hat\imath\), so only the component of area facing the x-direction matters:

\[\phi = E\,A_{\perp}, \qquad A_{\perp} = A\cos\theta.\]

Step 2: Part (a). A square parallel to the yz-plane is fully face-on to the x-directed field, so its whole area projects: \(A_{\perp} = A = 1\times10^{-2}\,\text{m}^{2}\).

\[\phi_a = (3\times10^{3})(1\times10^{-2}) = 30\,\text{N m}^{2}/\text{C}.\]

Step 3: Part (b). Tilting the normal by \(60^\circ\) shrinks the projected area to \(A\cos 60^\circ = (1\times10^{-2})(0.5) = 0.5\times10^{-2}\,\text{m}^{2}\).

\[\phi_b = (3\times10^{3})(0.5\times10^{-2}) = 15\,\text{N m}^{2}/\text{C}.\]

The tilted square intercepts exactly half as many field lines, so half the flux.

\[\boxed{\phi_a = 30, \quad \phi_b = 15 \ \text{N m}^{2}/\text{C}}\]