Answer:
The magnitude of the electric field is [tex]0.513\times10^{12}\ N/C[/tex].
Explanation:
Given that,
Radius [tex]r= 5.3\times10^{-11}\ m[/tex]
Speed [tex]v =2.2\times10^{6}\ m/s[/tex]
We need to calculate the electric field
Using formula of the electric field
[tex]E=\dfrac{kQ}{r^2}[/tex]
Where, Q = charge
r = radius
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times1.6\times10^{-19}}{(5.3\times10^{-11})^2}[/tex]
[tex]E=0.513\times10^{12}\ N/C[/tex]
Hence, The magnitude of the electric field is [tex]0.513\times10^{12}\ N/C[/tex].
The magnitude of the electric field is about 5.1 × 10¹¹ N/C
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Electric charge consists of two types i.e. positively electric charge and negatively electric charge.
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There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
[tex]\boxed {F = k \frac{Q_1Q_2}{R^2} }[/tex]
[tex]\boxed {E = F \div q = k \frac{Q}{R^2} }[/tex]
F = electric force (N)
E = electric field strength (N/C)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
radius of orbit = R = 5.3 × 10⁻¹¹ m
speed of electron = v = 2.2 × 10⁶ m/s
charge of electron = Q = 1.6 × 10⁻¹⁹ C
Asked:
magnitude of the electric field = E = ?
Solution:
[tex]E = k \frac{Q}{R^2}[/tex]
[tex]E = 9 \times 10^9 \times \frac{1.6 \times 10^{-19}}{(5.3 \times 10^{-11})^2}[/tex]
[tex]E \approx 5.1 \times 10^{11} \texttt{ N/C}[/tex]
[tex]\texttt{ }[/tex]
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Grade: High School
Subject: Physics
Chapter: Static Electricity
