Ask question

Ask question

All categories

3 years ago

13

A 2.00-gram air inflated balloon is given an excess negative charge, q1 = -3.25 × 10-8 C, by rubbing it with a blanket. It is fo

und that a charged rod can be held above the balloon at a distance of d = 5.00 cm to make the balloon float. In order for this to occur, what polarity of charge must the rod possess? How much charge, q2, does the rod have? Assume the balloon and rod to be point charges. The Coulomb force constant is 1/(4π ε0) = 8.99 × 109 N·m2/C2 and the acceleration due to gravity g = 9.81 m/s2.

1 answer:

Llana [10]3 years ago

3 0

Answer:

The rod`s charge must be positive, because the gravity force is pointing downwards and the electrostatic force must be pointing upwards (in order to balance the gravity force)

The charge is q_2 = 1.667 times 10^(-7) C

Explanation:

F_e = F_g

where F_g = m g and F_e= (1/4 pi e_0)*(q_1*q_2)/d^2,

please see the file attached for more details.

You might be interested in

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

If the sun exploded would the solar system collaspe?

Sav [38]

Answer:

Yes, yes it would since we need light

Explanation:

5 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

3 years ago

The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass

Triss [41]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

spring constant of the spring attached to the input piston, $k=1600\ N.m^{-1}$

mass subjected to the output plunger, $m=40\ kg$

<u>Now, the force due to the mass:</u>

$F=m.g$

$F=40\times 9.8$

$F=392\ N$

<u>Compression in Spring:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{392}{1600}$

$\Delta x=0.245\ m$

or

$\Delta x=245\ mm$

8 0

3 years ago

How are hot spots used to track plate motion

wel

Hot spots are fixed locations in the earth's mantle where heat from the earth's interior rises to the surface and produces volcanism. the earth's plates, which are slowly but constantly moving, are pierced by the uprising magma. as they move away form the hotspot, the volcanoes become dormant and are replaces by new volcanoes. The direction of the line formed from the previous volcanoes indicates the direction :)))

7 0

3 years ago

Read 2 more answers