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2 years ago

Explain the phenomenon of polarization when a dielectric slab is subjected to an electric field?

1 answer:

8 0

A dielectric, insulating material, or an extremely bad conductor of electrical current. Due to the absence of loosely bound, or free, electrons that could wander through the material, unlike metals, dielectrics practically do not conduct current when exposed to an electric field. Electric polarization takes place instead.

<h3>What is an Electric field?</h3>

An electric field is an electrical property associated with every point in the space of any form of charge. An electric field is also described as the electric force per unit charge.
Variable magnetic fields or electric charges are frequently the cause of electric fields. Volts per meter, a unit used in the SI, express electric field strength.
The force acting on the positive charge is assumed to be exerted in the direction of the field. The electric field is directed radially inwards toward the negative point charge and radially outwards from the positive charge.
Electric charge or magnetic fields with variable amplitudes can produce an electric field. The attraction forces that keep together atomic nuclei and electrons at the atomic scale are brought on by the electric field.

The phenomenon of polarization when a dielectric slab is subjected to an electric field:

To learn more about the electric field, refer to:

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You and a friend are studying late at night. There are three 110 W light bulbs and a radio with an internal resistance of 56.0 Ω

Sliva [168]

Answer:

The total electrical power we are using is: 1316 W.

Explanation:

Using the ohm´s law $V=I*R$ and the formula for calculate the electrical power, we can find the total electrical power that we are using. First we need to find each electrical power that is using every single component, so the radio power is: $I=\frac{V}{R}=\frac{110 (v)}{56(ohms)}=1.96(A)$ , so the radio power is: $P=I*V=1.96(A)*110(v)=216(W)$ , then we find the pop-corn machine power as: $P=I*V=7(A)*110(v)=770(W)$ and finally there are three light bulbs of 110(W) so: P=3*110(W)=330(W) and the total electrical power is the adding up every single power so that: P=330(W)+770(W)+216(W)=1316(W).

8 0

3 years ago

RP 1 - Specific Heat Capacity GCSE Exam Questions (B.S.G)

Stels [109]

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.

We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.

$Q = c \times m \times \Delta T = \frac{4200J}{kg.\° C} \times 0.50kg \times (-22\° C) \times \frac{1kJ}{1000J} = -46.2 kJ$

where,

c: specific heat capacity of water
m: mass of water
ΔT: change in the temperature

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

Learn more: brainly.com/question/16104165

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2 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

3 0

3 years ago

!!!!!!!!!!!PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!

Alexeev081 [22]

You might want to do a re-erp on this equation

5 0

3 years ago

Read 2 more answers

In Ampere's law, ∮????⃗∙????????⃗=????0???? the direction of the integration around the path:

kupik [55]

Answer:

C) must be such as to follow the magnetic field lines.

Explanation:

Ampere's circuital law helps us to calculate magnetic field due to a current carrying conductor. Magnetic field due to a current forms closed loop around the current . If a net current of value I creates a magnetic field B around it , the line integral of magnetic field around a closed path becomes equal to μ₀ times the net current . It is Ampere's circuital law . There may be more than one current passing through the area enclosed by closed curve . In that case we will take net current by adding or subtracting them according to their direction.

It is expressed as follows

∫ B.dl = μ₀ I . Here integration is carried over closed path . It may not be circular in shape. The limit of this integration must follow magnetic field lines.

the term ∫ B.dl is called line integral of magnetic field.

3 0

3 years ago