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2 years ago

Difference between discrete and continuous charge distribution?

2 answers:

5 0

Discrete systems are those systems in which are made up of finite component particles a which are non-homogeneously arranged such that no smooth variation exists. It is such that all constituent particles have properties which vary randomly. They are direct opposite to continuous systems, which are smooth arrangement of particles which cannot be individually taken into consideration.
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nadya68 [22]2 years ago

5 0

Interesting question. I think it's a question of scale. After all, charge is "discrete" or "quantised" into electronic charges.
So, relatively few of them - eg in an atom - would be discrete.
A lot of them - eg in an electric current - would be "continuous" (or an avalanche of discretes = a continuum ... )

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What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?

babunello [35]

Answer:

m³/(kg⋅s²)

Explanation:

Hello.

In this case, since the involved formula is:

$F=G*\frac{m_1m_2}{r^2}$

By writing a dimensional analysis with the proper algebra handling, we obtain:

$N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}$

Thus, answer is:

m³/(kg⋅s²)

Note that the [=] is used to indicate the units of G.

Best regards

4 0

2 years ago

A 1.97-pF capacitor is connected to a 9.0-V battery and fully charged. How many electrons did the battery transfer from one capa

Dovator [93]

Answer:

1.11×10⁸ Electrons.

Explanation:

Applying,

Q = CV..................... Equation 1

Where Q = Total charge on the capacitor, C = Capacitance of the capacitor, V = Voltage of the battery.

Given: C = 1.97 pF = 1.97×10⁻¹² F, V = 9.0 V

Substitute this value into equation 1

Q = 1.97×9×10⁻¹²

Q = 1.773×10⁻¹¹ C.

If the charge on one electron = 1.602×10⁻¹⁹ C.

Therefore number of electron = 1.773×10⁻¹¹ /1.602×10⁻¹⁹

Number of electron = 1.11×10⁸

8 0

2 years ago

Read 2 more answers

A student used apparatus as shown above. The beaker held 750 g of a liquid. The current from the power supply was 1.8 A and the

liq [111]

Answer:

$2365.71\ \text{J/kg}^{\circ}\text{C}$

Explanation:

V = Voltage = 230 V

I = Current = 1.8 A

$\Delta T$ = Temperature change = $(40-12)^{\circ}\text{C}$

t = Time taken = 2 minutes

m = Mass of liquid = 750 g

c = Specific heat capacity of the liquid

Energy required to heat the water is equal to the heat released due to current passing

$mc\Delta T=VIt\\\Rightarrow c=\dfrac{VIt}{m\Delta T}\\\Rightarrow c=\dfrac{230\times 1.8\times 2\times 60}{0.75\times (40-12)}\\\Rightarrow c=2365.71\ \text{J/kg}^{\circ}\text{C}$