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mash [69]
3 years ago
6

The usual electrosurgical apparatus contains an oscillator that produces a high-frequency current of frequencies above 500,000 c

ycles per second (500 kHz). Two electrodes are connected to the device. The large electrode is placed beneath the patient, and the "active" second electrode is sterilized before use. The purpose of the current is to raise the temperature of a small part of tissue so as to coagulate it, boil the tissue fluid, or char the tissue [4]. The machine may be used in different ways.
Listed below are some of these medical techniques:

ELECTROCOAGULATION: electrodes are in direct contact, and the heat in the tissues causes coagulation.
ELECTRODESICCATION: the active electrode is held a very short distance from the tissue. This causes a spark to jump from the electrode to the tissue, which in turn causes evaporation of moisture with coagulation, penetrating more deeply than when in contact.
ELECTROHEMOSTASIS: the bleeders are clamped with a hemostat in the usual manner. The electrode is touched to the clamp, and the dehydration and coagulation take place under the clamp.
ELECTRIC CUTTING: the fine tungsten wire makes a minute arc (a continuous spark or tiny flame) just before touching the tissue. This cuts the tissue and seals off the blood vessels. The arc separates the wire from the tissue so that the wire does not actually touch it. That is, the arc does the cutting rather than the wire.
Electrocautery is especially valuable in surgery of the brain, prostate gland, and the liver [4].

In which of the following medical techniques is the active electrode in direct contact with the tissue?

Group of answer choices

A、Electrodessication

B、Electrocoagulation

C、Electrohemostais

D、Electric cutting
Physics
1 answer:
AURORKA [14]3 years ago
8 0

Answer:

B Electrocoagulation.

It is a process that involves the use of high frequency current to cause coagulation usually of protiens in tissues and also the destruction of abnormal tissues making use of the heat generated by the electric current. This tissue could be a tumor or some other kind of lesion.

Explanation:

The passage of electric current through the electrode (second electrode that has been sterilized) in contact with the tissue causes the molecules within them to vibrate. This leads to a sharp increase in the temperature of the tissue due to the heat generated by the electric current. This is what causes the coagulation of the proteins within the tissue and also burns the tissue thereby effectively destroying it.

Electrocoagulation also called electrocautery, electrofulguration, and fulguration.

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Answer:

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

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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}

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