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Sergio039 [100]

2 years ago

4

A charge of 0.80nC is placed at the center of a cube that measures 4.0 m along each edge. What is the electric flux through one

face of the cube?. 1. 5 N m2/C
2. 90 N m2/C
3. 45 N m2/C
4. 64 N m2/C
5. 23 N m2/C

1 answer:

Nutka1998 [239]2 years ago

8 0

Answer:

Explanation:

Given

Charge of $Q=0.8\ nC$

Length of side of cube $a=4\ m$

Total Flux through the cube $\phi =\frac{Q}{\epsilon _0}$

$\epsilon _0=8.854\times 10^{-12}\ C^2/N-m^2$

total Flux $\phi=\frac{0.8\times 10^{-9}}{8.854\times 10^{-12}}$

$\phi =90.35\ N-m^2/C$

As cube has 6 faces , so flux through each face will be similar

thus flux through each face $\phi'=\frac{\phi }{6}$

$\phi '=\frac{90.35}{6}=15\ N-m^2/C$

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The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

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$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

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