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torisob [31]
3 years ago
14

A flat sheet if in the shape of a rectangle with sides of length 0.400 m and 0.600 m. The sheet is immersed in a uniform electri

c field of magnitude 95 N/C that is directed 20° from the plane of the sheet. Find the magnitude of the electric flux through the sheet.
Physics
1 answer:
Nataly [62]3 years ago
6 0

Answer:

Φ= 17 N•m²•C⁻¹

Explanation:

Gauss's Law states that electric flux equals the surface integral of E•dA. But since we are given all the variables as finite values, we can simplify it into EAcosφ.

-E is given as 95N/C

-A is simply (.4)(.6)=.24m²

-φ is the angle between the E field/vector and the normal/perpendicular vector to the surface. We know that E makes a 20° to the surface here, so the angle φ=(90-20)°=70°. So the E vector makes a 70° angle to the normal of the surface. (I can see this portion as being the point of confusion, as it was for me at first.)

With all that we can say that the flux Φ is:

Φ=(95)(0.24)(cos[70°])=17.4384... N•m²•C⁻¹

I'll approximate to 2 sigfigs in my answer, since that'd be the technical answer.

*I believe V/m are also correct units for electric flux.

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Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

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(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

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