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3 years ago

A free positive charge released in an electric field will _____

Physics

1 answer:

morpeh [17]3 years ago

4 0

Answer:

Accelerates in the same direction as the field

Explanation:

When a charge is immersed in an electric field, the charge experiences an electric force, whose magnitude is given by

F = qE

where q is the magnitude of the charge and E is the magnitude of the electric field.

The direction of the force depends on the sign of the charge. In particular,

- For a positive charge, F and E have the same direction

- For a negative charge, F and E have opposite directions

Therefore, a free positive charge will experience a force in the same direction as the field - therefore, it will accelerate in the same direction.

You might be interested in

Why do we call the microscope a compound light microscope

Ede4ka [16]

Explanation:

Types of light microscope

1. Compound , and 2. Stereo Microscope

Compound microscope has two lens system also called compound lens system. The objective lens and the eyepiece lens. The magnification provided by the objective lens is compounded by the eyepiece lens, the a higher magnification is observed.

6 0

3 years ago

a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra

Feliz [49]

Answer:

0.331m

Explanation:

wave equation: v=f λ

v=30000m/s

f=90500 Hz

λ= $\frac{v}{f}$

λ= $\frac{30000}{90500}$

λ= 0.311m

5 0

2 years ago

A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog

DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

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8 0

3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

For a wire has a circular cross section with a radius of 1.23mm.

Mila [183]

Answer:

$5.731\times 10^{-5}\ m/s$

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = Conduction electron density in copper = $8.49\times 10^{28}\ electrons/m^3$

$v_d$ = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

$I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s$

The drift speed of the electrons is $5.731\times 10^{-5}\ m/s$

$v_d=\dfrac{I}{neA}$

From the equation we can see the following

$v_d\propto \dfrac{1}{n}$

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0

3 years ago