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Since the test charge cancels, we see that
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 $q$ .
Calculate the strength and direction of the electric field $E$ due to a point charge of 2.00 nC (nano-Coulombs) 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
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 $Q$ .
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,
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.
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.
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.
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?
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 Problem-Solving 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) $4\text{.}\text{80}\times {\text{10}}^{-\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}(\text{east})$
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