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3 years ago

What is the difference between centrifugal force and centripetal force???

1 answer:

5 0

Centripetal force is defined as, "the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation," while centrifugal force is defined as "the apparent force that is felt by an object moving in a curved path that acts outwardly away from the center of rotation," according to Merriam Webster Dictionary.

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Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o

Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

qA = ½ q

qB = ½ q

qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

qA = 1/2 (q / 2) = ¼ q

qC = ¼ q

qB = ½ q

At this point all spheres have a charge,

qA = ¼ q

qb = ½ q

qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

Q = ½ q + ¼ q = ¾ q

When splitting the charge

qB = ½ ¾ q = 3/8 q

qC = ½ ¾ q = 3/8 q

qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

Q = qA + QB + QC

Q = ½ q + 3/8 q + ¼ q

Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0

3 years ago

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

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

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$