Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 cm. (a) W

hat is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

Physics

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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that

serg [3582]

Response:

Reasoning:

We will utilize a Gaussian surface that resembles the curved wall of a cylinder, with a radius of 3mm and a length of 1 unit directed parallel to the wire axis.

The charge within this cylinder amounts to 250 x 10⁻⁹ C.

Let E denote the electric field at the curved surface, perpendicular to it.

The total electric flux leaving the curved surface

is calculated as 2π r x 1 x E

or 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's law, the total flux is given by the charge within divided by ε (the charge inside the cylinder being 250 x 10⁻⁹C)

equals 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹² (where ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹²)

resulting in 11.3 x 10³ weber.

Thus,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E = 11.3 x 10³ / 2 x 3.14 x 3 x 10⁻³

=.599 x 10⁶ N /C.

4 0

1 month ago

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

inna [3103]

Answer:

The ratio of mass that is discarded is determined by this equation:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

The force acting on an object in motion is defined by the equation:

F = ma

Additionally, there is a gravitational force consistently acting downwards on the object, defined as g = 9.8 ms⁻²

For convenience, we will utilize a positive notation for downward acceleration and a negative notation for upward acceleration.

Case 1:

The hot air balloon has mass = M

Acceleration = a

Upward thrust from hot air = F = constant

Gravitational force acting downward = Mg

The net force on the balloon can be expressed as:

Ma = Gravitational force - Upward Force

Ma = Mg - F (since the balloon moves downward, that means Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After releasing the ballast, the new mass becomes m. The new upward acceleration is -a/2:

The net force is expressed as:

-m(a/2) = mg - F (The balloon is moving upwards, hence F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Determining the fraction of the mass initially dropped:

$M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]$