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1 year ago

6

What two things must exist for an electric current to be produced?A pair of highly conductive materials and a conducting path be

tween themA source of voltage and an open circuitA negatively charged electrode and a positively charged electrodeAn electric potential between two bodies and a conducting path joining the bodies

1 answer:

Andrei [34K]1 year ago

5 0

The two things that must exist for an electric current to be produced are:

An electric potential between two bodies and a conducting path joining the bodies.

This comes from the fact that when there is a potential difference there is an electrical field and hence a force that makes some free charges to move in the conductor. For this reason, there's an electrical current.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

4 years ago

Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc

Vanyuwa [196]

Answer:

$1\cdot 10^{-8} N$ to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

$q_1=2nC=2\cdot 10^{-9}C$ is the magnitude of charge 1

$q_2=5nC =5\cdot 10^{-9}C$ is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

$F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N$

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0

3 years ago

A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is

Snezhnost [94]

Answer:

La frecuencia será la misma en los dos medios, y en el vacio, no varia.

b) Vacío : n = 3,684 10 m 1,33 = 4 900 Å ,3161 10 m 55,1 n

,6 105 10 Hz

3,161 10 m

v 1,93 10 m/s 193548,38 km/s =

1,55

10 3 m/s

n

c

Vidrio v: =

,6 107 10 Hz

3,684 10 m

v 2,25 10 m/s 225 000km/s =

33,1

10 3 m/s

n

c

a

6 0

3 years ago

A bicycle-cab driver takes two customers across the park. The driver exerts a constant force of 1,400 Newtons as he pedals the c

seraphim [82]

Power is the work performed divided by time in seconds. The work is the force times distance:
$W=Fd=1400*200=280000J$

the time in seconds is 60*11+30=690s so the power is

$P=\frac{280000J }{690s}=405.8W$

So choice B

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3 years ago

What is the definition of speed

lana [24]

The rate at which someone or something can move or operate very fast.

4 0

4 years ago

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