Question

2025 - 2026 - 1 exercise for chapter 22
student id no. s11a100601 name abigail odusu
(1) the electric field in a region of space is oriented along the positive y - axis. a circle of radius r is placed in the x - plane. the flux of the electric field through this circle is φ. the same electric field passing through a second circle of radius 2r parallel to x - plane would result in a flux equal to (
(a) φ
(b) 4φ
(c) 2φ
(d) 3φ
(2) if a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it. what has been the angle of rotation? (
(a) 30°
(b) 45°
(c) 60°
(d) 90°
(3) a region of space contains a uniform electric field oriented along the y - axis. a frame of surface area a is placed perpendicular to the y - axis in the x - plane. the magnitude of the electric flux through this frame is φ₀. a second frame is placed in the same electric field in such a way that the magnitude of the electric flux through it is φ₀/2. how is the plane of second frame oriented with respect to the plane of the first one?
(a) at a 30° angle
(b) at a 60° angle
(c) parallel to the first frame
(d) perpendicular to the first frame

Explanation:

Step1: Recall electric - flux formula

The electric - flux through a surface is given by $\varPhi = E\cdot A\cdot\cos\theta$, where $E$ is the electric - field magnitude, $A$ is the area of the surface, and $\theta$ is the angle between the electric - field vector and the normal to the surface. For a circle in the $x - y$ plane with an electric field along the $y$ - axis, $\theta = 0$ and $\varPhi=E\cdot A$. The area of a circle is $A=\pi r^{2}$.

Step2: Calculate the areas of the two circles

Let the radius of the first circle be $R$, so its area $A_1=\pi R^{2}$ and the flux $\varPhi_1 = E\cdot A_1=E\cdot\pi R^{2}=\varPhi$. The radius of the second circle is $2R$, and its area $A_2=\pi(2R)^{2}=4\pi R^{2}$. Then the flux $\varPhi_2 = E\cdot A_2=E\cdot4\pi R^{2}=4\varPhi$.

Step3: Solve for the rotation angle in the second question

The maximum electric flux through a surface in a uniform electric field occurs when $\theta = 0$ and $\varPhi_{max}=E\cdot A$. When the flux is half of the maximum value, $\varPhi=\frac{1}{2}\varPhi_{max}$. Using the formula $\varPhi = E\cdot A\cdot\cos\theta$, we have $\frac{1}{2}E\cdot A=E\cdot A\cdot\cos\theta$. Solving for $\theta$, we get $\cos\theta=\frac{1}{2}$, so $\theta = 60^{\circ}$.

Step4: Solve for the orientation of the second frame in the third question

The electric flux through a surface is $\varPhi = E\cdot A\cdot\cos\theta$. Let the area of the first frame be $A_1 = A$ and its flux $\varPhi_1=\varPhi_0 = E\cdot A$. For the second frame, $\varPhi_2=\frac{1}{2}\varPhi_0$. Using $\varPhi_2 = E\cdot A\cdot\cos\theta$, and since $\varPhi_2=\frac{1}{2}\varPhi_0=\frac{1}{2}E\cdot A$, we have $\cos\theta=\frac{1}{2}$, so $\theta = 60^{\circ}$.

Answer:

(1) B. $4\varPhi$
(2) C. $60^{\circ}$
(3) B. at a $60^{\circ}$ angle