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Chapter 17: Q7Q (page 473)

If \({\bf{V = 0}}\) at a point in space, must \({\bf{\vec E = 0}}\) ? If \({\bf{\vec E = 0}}\) at some point, must \({\bf{V = 0}}\) at that point? Explain. Give examples for each.

Short Answer

Expert verified

When\(V = 0\)at a point in space, the electric field must be zero,\({\rm{\vec E}} = 0\).

No, it is not mandatory for \(V = 0\) when \({\rm{\vec E}} = 0\).

Step by step solution

01

Understanding the electric field

The electric field vector is the negative gradient of the electric potential at any point.

The electric field at any point is

\(\overrightarrow {\rm{E}} = - \overrightarrow \nabla V\).

02

Evaluation of electric field and electric potential

The electric field is zero. Then from equation (i),

\(\begin{aligned}{c}\vec \nabla V &= 0\\V = {\rm{constant}}\end{aligned}\)

The electric potential is zero or any constant value; then, the electric field is zero at that point.

So when\(V = 0\)at a point in space, the electric field must be zero,\({\rm{\vec E}} = 0\).

When\({\rm{\vec E}} = 0\), the value of electric potential is either zero or constant.

Thus, it is not mandatory for \(V = 0\) when \({\rm{\vec E}} = 0\).

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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 \(\left( {{\bf{1}}\;{\bf{fF = 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{F}}} \right)\) 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 \({\bf{0}}{\bf{.30}}\;{\bf{fC/s}}\). 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.

A parallel-plate capacitor with plate area\({\bf{A = 2}}{\bf{.0}}\;{{\bf{m}}{\bf{2}}}\)and plate separation\({\bf{d = 3}}{\bf{.0}}\;{\bf{mm}}\)is connected to a 35-V battery (Fig. 17–51a).

(a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor.

(b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength K =3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17–51b). What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor?\(\left( {{\bf{1}}\;{\bf{byte = 8}}\;{\bf{bits}}{\bf{.}}} \right)\)

FIGURE 17-51 Problem 96

(II) Draw a conductor in the oblong shape of a football. This conductor carries a net negative charge -Q, Draw in a dozen or so electric field lines and equipotential lines.

(II) An electric field greater than about \({\bf{3 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{V/m}}\)causes 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?

A proton \(\left( {{\bf{Q = + e}}} \right)\) and an electron \(\left( {{\bf{Q = - e}}} \right)\) are in a constant electric field created by oppositely charged plates. You release the proton from near the positive plate and the electron from near the negative plate. Which feels the larger electric force?

(a) The proton.

(b) The electron.

(c) Neither—there is no force.

(d) The magnitude of the force is the same for both and in the same direction.

(e) The magnitude of the force is the same for both but in opposite directions.
See all solutions

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