Question

What shaped Gaussian surface do you build around a point charge? Why? A uniform electric field of 500 N/C has a direction of 30 degrees from the ground Calculate the flux through the cushion of your seat as you drive through this field Assume your seat is 0.5 m by 0.7 m and is parallel to the ground Using Gauss' law, calculate the electric field magnitude at a distance of 12.5 cm from a charged sphere with a charge of 3.33e-5 C.

Answer 1

Part a)

Gaussian Surface must be spherical

Part b)

`\phi = 87.5 Nm^2/C`

Part c)

`E = 1.92 * 10^7 N/C`

Step-by-step explanation:

Part a)

We can build a spherical Gaussian surface around a point charge

Because the flux of electric field around the spherical Gaussian surface will be uniform throughout the surface.

Part b)

Area of the cushion is given as

`A = 0.5 * 0.7`

`A = 0.35 m^2`

now electric field has two components

`E_x = 500 cos30 = 433 N/C`

`E_y = 500 sin30 = 250 N/C`

now we know that flux due to horizontal component of field is zero

while flux due to vertical component is given as

`\phi = E_y .A`

`\phi = (250)(0.35)`

`\phi = 87.5 Nm^2/C`

Part c)

Electric flux due to a spherical surface is given as

`E. A = (q)/(\epsilon_0)`

`E(4\pi R^2) = (3.33 * 10^(-5))/(8.85 * 10^(-12))`

`E = 1.92 * 10^7 N/C`