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 Electric field: concept of a
$E=\left\frac{F}{q}\right=k\frac{\text{qQ}}{{\mathrm{qr}}^{2}}=k\frac{\leftQ\right}{{r}^{2}}.$
Since the test charge cancels, we see that
$E=k\frac{\leftQ\right}{{r}^{2}}.$
The electric field is thus seen to depend only on the charge
$Q$ and the distance
$r$ ; it is completely independent of the test charge
$q$ .
Calculating the electric field of a point charge
Calculate the strength and direction of the electric field
$E$ due to a point charge of 2.00 nC (nanoCoulombs) at a distance of 5.00 mm from the charge.
Strategy
We can find the electric field created by a point charge by using the equation
$E=\text{kQ}/{r}^{2}$ .
Solution
Here
$Q=2\text{.}\text{00}\times {\text{10}}^{9}$ C and
$r=5\text{.}\text{00}\times {\text{10}}^{3}$ m. Entering those values into the above equation gives
$\begin{array}{lll}E& =& k\frac{Q}{{r}^{2}}\\ & =& (\text{8.99}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/C}}^{2})\times \frac{(\text{2.00}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{C})}{(\text{5.00}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}}\\ & =& \text{7.19}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C.}\end{array}$
Discussion
This
electric field strength is the same at any point 5.00 mm away from the charge
$Q$ that creates the field. It is positive, meaning that it has a direction pointing away from the charge
$Q$ .
Calculating the force exerted on a point charge by an electric field
What force does the electric field found in the previous example exert on a point charge of
$\mathrm{\u20130.250}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ ?
Strategy
Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field
$\mathbf{\text{E}}=\mathbf{\text{F}}/q$ rearranged to
$\mathbf{\text{F}}=q\mathbf{\text{E}}$ .
Solution
The magnitude of the force on a charge
$q=0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{\mu C}$ exerted by a field of strength
$E=7\text{.}\text{20}\times {\text{10}}^{5}$ N/C is thus,
$\begin{array}{lll}F& =& \text{qE}\\ & =& (\text{0.250}\times {\text{10}}^{\text{\u20136}}\phantom{\rule{0.25em}{0ex}}\text{C})(7.20\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N/C})\\ & =& \text{0.180 N.}\end{array}$
Because
$q$ is negative, the force is directed opposite to the direction of the field.
Discussion
The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.
Section summary
 The electrostatic force field surrounding a charged object extends out into space in all directions.
 The electrostatic force exerted by a point charge on a test charge at a distance
$r$ depends on the charge of both charges, as well as the distance between the two.
 The electric field
$\mathbf{\text{E}}$ is defined to be
$\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q,}$
where
$\mathbf{\text{F}}$ is the Coulomb or electrostatic force exerted on a small positive test charge
$q$ .
$\mathbf{\text{E}}$ has units of N/C.
 The magnitude of the electric field
$\mathbf{\text{E}}$ created by a point charge
$Q$ is
$\mathbf{\text{E}}=k\frac{\leftQ\right}{{r}^{2}}.$
where
$r$ is the distance from
$Q$ . The electric field
$\mathbf{\text{E}}$ is a vector and fields due to multiple charges add like vectors.
Conceptual questions
Why must the test charge
$q$ in the definition of the electric field be vanishingly small?
Are the direction and magnitude of the Coulomb force unique at a given point in space? What about the electric field?
Problem exercises
What is the magnitude and direction of an electric field that exerts a
$2\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$ upward force on a
$\mathrm{\u20131.75}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge?
What is the magnitude and direction of the force exerted on a
$3.50\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge by a 250 N/C electric field that points due east?
$8\text{.}\text{75}\times {\text{10}}^{4}$ N
Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).
(a) What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?
(a)
$6\text{.}\text{94}\times {\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{C}$
(b)
$6\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{N/C}$
Calculate the initial (from rest) acceleration of a proton in a
$5\text{.}\text{00}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N/C}$ electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the ProblemSolving Strategy for electrostatics.
(a) Find the direction and magnitude of an electric field that exerts a
$4\text{.}\text{80}\times {\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}$ westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?
(a)
$\text{300}\phantom{\rule{0.25em}{0ex}}\text{N/C}\phantom{\rule{0.25em}{0ex}}(\text{east})$
(b)
$4\text{.}\text{80}\times {\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}(\text{east})$
Questions & Answers
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is this allso about nanoscale material
Almas
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
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William
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nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
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Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
Application of nanotechnology in medicine
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Kamaluddeen
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ya I also want to know the raman spectra
Bhagvanji
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Damian
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Professor
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Alexandre
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Alexandre
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Rafiq
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Damian
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What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
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Rafiq
Rafiq
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Rafiq
Rafiq
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Anam
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Hafiz
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Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:
OpenStax, Concepts of physics with linear momentum. OpenStax CNX. Aug 11, 2016 Download for free at http://legacy.cnx.org/content/col11960/1.9
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