7.3 Calculations of electric potential  (Page 5/7)

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Check Your Understanding What is the potential on the axis of a nonuniform ring of charge, where the charge density is $\lambda \left(\theta \right)=\lambda \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta$ ?

It will be zero, as at all points on the axis, there are equal and opposite charges equidistant from the point of interest. Note that this distribution will, in fact, have a dipole moment.

Summary

• Electric potential is a scalar whereas electric field is a vector.
• Addition of voltages as numbers gives the voltage due to a combination of point charges, allowing us to use the principle of superposition: ${V}_{P}=k\sum _{1}^{N}\frac{{q}_{i}}{{r}_{i}}$ .
• An electric dipole consists of two equal and opposite charges a fixed distance apart, with a dipole moment $\stackrel{\to }{\text{p}}=q\stackrel{\to }{\text{d}}$ .
• Continuous charge distributions may be calculated with ${V}_{P}=k\int \frac{dq}{r}$ .

Conceptual questions

Compare the electric dipole moments of charges $±Q$ separated by a distance d and charges $±Q\text{/}2$ separated by a distance d /2.

The second has 1/4 the dipole moment of the first.

Would Gauss’s law be helpful for determining the electric field of a dipole? Why?

In what region of space is the potential due to a uniformly charged sphere the same as that of a point charge? In what region does it differ from that of a point charge?

The region outside of the sphere will have a potential indistinguishable from a point charge; the interior of the sphere will have a different potential.

Can the potential of a nonuniformly charged sphere be the same as that of a point charge? Explain.

Problems

A 0.500-cm-diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0-pC charge on its surface. What is the potential near its surface?

$V=144\phantom{\rule{0.2em}{0ex}}\text{V}$

How far from a $1.00\text{-}\mu \text{C}$ point charge is the potential 100 V? At what distance is it $2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{V?}$

If the potential due to a point charge is $5.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{V}$ at a distance of 15.0 m, what are the sign and magnitude of the charge?

$V=\frac{kQ}{r}\to Q=8.33\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{C}$ ;
The charge is positive because the potential is positive.

In nuclear fission, a nucleus splits roughly in half. (a) What is the potential $2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{m}$ from a fragment that has 46 protons in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance?

A research Van de Graaff generator has a 2.00-m-diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface? (b) At what distance from its center is the potential 1.00 MV? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?

a. $V=45.0\phantom{\rule{0.2em}{0ex}}\text{MV}$ ;
b. $V=\frac{kQ}{r}\to r=45.0\phantom{\rule{0.2em}{0ex}}\text{m}$ ;
c. $\text{Δ}U=132\phantom{\rule{0.2em}{0ex}}\text{MeV}$

An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a grounded object.

(a) What charge is on the sphere? (b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?

(a) What is the potential between two points situated 10 cm and 20 cm from a $3.0\text{-}\mu \text{C}$ point charge? (b) To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?

$V=kQ/r$ ; a. Relative to origin, find the potential at each point and then calculate the difference.
$\text{Δ}V=135\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{V}$ ;
b. To double the potential difference, move the point from 20 cm to infinity; the potential at 20 cm is halfway between zero and that at 10 cm.

Find the potential at points ${P}_{1},{P}_{2},{P}_{3},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{P}_{4}$ in the diagram due to the two given charges. Two charges $–2.0\phantom{\rule{0.2em}{0ex}}µ\text{C}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{+}2.0\phantom{\rule{0.2em}{0ex}}µ\text{C}$ are separated by 4.0 cm on the z -axis symmetrically about origin, with the positive one uppermost. Two space points of interest ${P}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{P}_{2}$ are located 3.0 cm and 30 cm from origin at an angle $30\text{°}$ with respect to the z -axis. Evaluate electric potentials at ${P}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{P}_{2}$ in two ways: (a) Using the exact formula for point charges, and (b) using the approximate dipole potential formula.

a. ${V}_{P1}=7.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{V}$
and ${V}_{P2}=6.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{V}$ ;
b. ${V}_{P1}=6.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{V}$ and ${V}_{P2}=6.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{V}$

(a) Plot the potential of a uniformly charged 1-m rod with 1 C/m charge as a function of the perpendicular distance from the center. Draw your graph from $\text{s}=0.1\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\text{s}=1.0\phantom{\rule{0.2em}{0ex}}\text{m}$ . (b) On the same graph, plot the potential of a point charge with a 1-C charge at the origin. (c) Which potential is stronger near the rod? (d) What happens to the difference as the distance increases? Interpret your result.

What is differential form of Gauss's law?
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a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
kya ye b.sc ka hai agar haa to konsa part
what is charge quantization
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
determine absolute zero
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
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this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
what is the weight of the earth in space
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
every law is important, but first law is conservation of energy, this state is the basic in physics, in this case first law is more important than other laws..
Mehmet
First Law describes o energy is changed from one form to another but not destroyed, but that second Law talk about entropy of a system increasing gradually
Mayowa
first law describes not destroyer energy to changed the form, but second law describes the fluid drection that is entropy. in this case first law is more basic accorging to me...
Mehmet
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
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pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe     By     By