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(II) An electric field greater than about 3×106V/mcauses air to break down (electrons are removed from the atoms and then recombine, emitting light). See Section 17–2 andTable 17–3. If you shuffle along a carpet and then reach for a doorknob, a spark flies across a gap you estimate to be 1 mm between your finger and the doorknob. Estimate the voltage between your finger and the doorknob. Why is no harm done?

Short Answer

Expert verified

The voltage between the finger and the doorknob is 3000 V. No harm is done because the amount of charge transferred is very small.

Step by step solution

01

Relation between Electric Field and Electric Potential

The electric potential energy per unit charge at any point in space is termed as the electric potential difference.

The potential difference (Vba)between two points a and bseparated by a distance d apart in a region of uniform electric field is related to the electric field (E) by the following relation:

E=Vbad=ΔVd … (i)

Here, negative sign indicates that the direction of electric field is in the direction of decreasing potential.

02

Given information

The electric field generated between the finger and the doorknob is approximately, E=3×106V/m

The distance between the finger and the doorknob is, d=1mm=1×103m

03

Determination of the voltage between the finger and the doorknob

From equation (i), the expression for magnitude of electric field can be written as:

E=ΔVd

So, the potential difference between finger and doorknob is,

ΔV=E.d=(3×106V/m)(1×103m)=3×103V=3000V

Thus, the voltage between the finger and the doorknob is 3000 V.

The charge transferred between the finger and the doorknob is very small. Therefore, no harm is done.

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Most popular questions from this chapter

In the dynamic random access memory (DRAM) of a computer, each memory cell contains a capacitor for charge storage. Each of these cells represents a single binary bit value of “1” when its 35-fF capacitor (1fF=1015F) is charged at 1.5 V, or “0” when uncharged at 0 V.

(a) When fully charged, how many excess electrons are on a cell capacitor’s negative plate?

(b) After charge has been placed on a cell capacitor’s plate, it slowly “leaks” off at a rate of about 0.30fC/s. How long does it take for the potential difference across this capacitor to decrease by 2.0% from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is “refreshed” many times per second.) Note: A DRAM cell is shown in Fig. 21–29.

If it takes an amount of work W to move two +q point charges from infinity to a distance d apart from each other, then how much work should it take to move three +q point charges from infinity to a distance d apart from each other?

(a) 2W.

(b) 3W.

(c) 4W.

(d) 6W.

(III) How much voltage must be used to accelerate a proton (radius 1.2×1015m) so that it has sufficient energy to just “touch” a silicon nucleus? A silicon nucleus has a charge of +14e, and its radius is about 3.6×1015m. Assume the potential is that for point charges.

Question: (II) A homemade capacitor is assembled by placing two 9-in. pie pans 4 cm apart and connecting them to the opposite terminals of a 9-V battery. Estimate (a) the capacitance, (b) the charge on each plate, (c) the electric field halfway between the plates, and (d) the work done by the battery to charge them. (e) Which of the above values change if a dielectric is inserted?

(II) The charge on a capacitor increases by 15 uC when the voltage across it increases from 97 V to 121 V. What is the capacitance of the capacitor?

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