Question

When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25 N .m2 /C. When the paper is turned 25° with respect to the field the flux through it is:_____.1. 02. 12 N .m2 /C3. 21 N .m2 /C4. 23 N .m2 /C5. 25 N .m2 /CB. A charged point particle is placed at the center of a spherical Gaussian surface. The electric flux is changed if:______.1. the sphere is replaced by a cube of the same volume2. the sphere is replaced by a cube of one-tenth the volume3. the point charge is moved off center (but still inside the original sphere)4. the point charge is moved to just outside the sphere5. a second point charge is placed just outside the sphereC. A charge of 0.8 × 10−9 C is placed at the center of a cube that measures 5 m along each edge. What is the electric flux through any 2 faces of the cube?

Answer 1

1) 23 N.m²/C

2) The point charge is moved just outside the sphere

3) 30 N.m²/C

Step-by-step explanation:

1) The flux through the field can be gotten by finding the dot product of the angle the later is turned, and the given uniform electric field.

25 * cos 25 =

25 * 0.9063 =

22.66 or approximately, 23 N.m²/C

2) The point charge is moved just outside the sphere

3) Using Gauss' Law, the electric flux through one face of the cube is given as Φ = Q / 6ε₀, thus,

Φ = 0.8*10^-9 / 6 * 8.85*10^-12

Φ = 0.8*10^-9 / 5.311*10^-11

Φ = 15 N.m²/C

And therefore, the flux through any 2 faces of the cube is 2 * 15 = 30 N.m²/C