Which of the following best explains why lightning occurs?

two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha

user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q = 4.0 μC = 4.0 × 10 C and
r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q' = -4.0 μC = -4.0 × 10 C and
r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

Learn more about electric potential here:

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3 0

2 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

wariber [46]

<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $\frac{11.7}{18}$

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0

4 years ago

Read 2 more answers

Help me answer this question and get yourself some points!

Marrrta [24]

Answer:

A×B=C×D

500×0.5=250×X

250=250×X

X=250/250=1

X=1 m

Explanation:

note: if the force plus two, the distance will be half.

4 0

3 years ago

Velocity has two pieces of information.what are they

seraphim [82]

They are speed and direction.

3 0

4 years ago

An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic

shepuryov [24]

Answer:

$v=\sqrt{\frac{kZe^2}{mr}}$

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

$F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}$ (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

$F=m\frac{v^2}{r}$ (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

$k\frac{Ze^2}{r^2}=m\frac{v^2}{r}$

And solving for v, we find an expression for the speed of the electron:

$v=\sqrt{\frac{kZe^2}{mr}}$

6 0

3 years ago