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vagabundo [1.1K]

3 years ago

12

The electric potential difference between the ground and a cloud in a particular thunderstorm is 2.5 × 109 V. What is the magnit

ude of the change in the electric potential energy of an electron that moves between the ground and the cloud?

1 answer:

muminat3 years ago

6 0

Answer:

U = - 4 x $10^{-10}$

Explanation:

ΔV = potential difference = $2.5 * 10^9$ Volts

q = charge on electron = $-1.6 * 10^{-19}$ C

electric potential energy is given as

U = q ΔV = ( $-1.6 * 10^{-19}$ ) ( $2.5 * 10^9$ )

= - 4 x $10^{-10}$

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3 years ago

Which of the following intermolecular forces explains why iodine (I2) is a solid at room temperature?

egoroff_w [7]

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3 0

4 years ago

How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners

Natasha_Volkova [10]

Answer:

$Potential\ Energy=Work \ Done=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V= $\frac{kq}{r}$

So,

$Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}$

Where:

k is Coulomb Constant= $8.99*10^9 Nm^2/C^2$

q is the charge on electron= $-1.6*10^-19 C$

r is the distance= $3.0*10^{-10}m$

For 3 Electrons Potential Energy or work Done is:

$Potential\ Energy=3*\frac{kq^2}{r}$

$Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

7 0

3 years ago

A 600 kg car is at test and then accelerated to 5m/s , what is its original kinetic energy

ycow [4]

Answer:

0 J

Explanation:

Kinetic energy is defined as:

KE = 1/2 m v²

where m is mass and v is velocity.

The car starts at rest, so it has zero velocity. Therefore, its initial kinetic energy is 0 J.

3 0

3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

4 years ago