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notsponge [240]

2 years ago

7

An electric dipole consists of a particle with a charge of 6 x 10–6 c at the origin and a particle with a charge of –6 x 10–6 c

on the x axis at x = 3 x 10–3 m. its dipole moment is:______.

1 answer:

mote1985 [20]2 years ago

8 0

An electric dipole consists of a particle with a charge of 6 x 10⁻⁶ c at the origin and a particle with a charge of –6 x 10⁻⁶ c on the x axis at x = 3 x 10⁻³ m. Its dipole moment is 18 x 10⁻⁹ Cm

Dipole moment of a dipole is dependent on the charge of the dipole and the distance between the two charges.

Electric Dipole consists of two charges which are equal and opposite in charge i.e. positive and negative charges.

Given,

Dipole moment, p = ?

Charge, q = 6 x 10⁻⁶C

Distance between charges, d = 3 x 10⁻³ m

Dipole moment (p) is given by:

p = charge x distance between the two charges

p = 6 x 10⁻⁶ x 3 x 10⁻³ Cm

p = 18 x 10⁻⁹ Cm

The dipole moment for the given charge configuration is 18 x 10⁻⁹ Cm

Learn more about Dipole moment here, brainly.com/question/14058533

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A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

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c)

In conclusion, the direction of the induced current will be Counterclockwise.

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4 0

2 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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3 years ago

A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-

kati45 [8]

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

$\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}$

Here,

$\Delta L$ = is the distance moved by the mirror M

$\lambda$ is the wavelenght of the light used.

$\Delta L$ = 0.017m

$\lambda = 656.45*10^{-9}m$

$\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}$

$\Delta m = 51793.72$

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

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3 years ago