Have you ever asked yourself why you stick a balloon to your hair after rubbing or why there is a flash all over the sky during a thunderstorm? In fact, all of these interesting events are because of Electric Force. In this article, we will discuss what is electric force, its S.I. unit, and calculation by using Coulomb’s law, with some real-life examples.
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Electric Force Definition:
Forces that results from an interaction of electric charges that can be either attractive or repulsive is called as an Electric Force.
Simply we can say that Electric force is the force exerted between two electrically charged objects. This force is a vector quantity, meaning it has both magnitude and direction. The direction of the electric force depends on whether the charges are alike (repulsive force) or different (attractive force).
The SI unit of electric force is the Newton (N).
Commonly Asked Questions
The electric force is described by Coulomb's Law, which states that the magnitude of the force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by:
$$
F=k_e \frac{\left|q_1 q_2\right|}{r^2}
$$
where:
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Electric Force is calculated by using Coulomb's law as stated above.
1. Identify the charges: Determine the values of $q_1$ and $q_2$.
2. Find the distance: Find the distance $r$ between the charges.
3. Put in the values: Substitute the values of $q_1, q_2, r$, and $k_e$ into the formula.
4. Solve for $F$ : Perform the calculations to find the electric force.
Commonly Asked Questions
The following are the electric force examples
Force that occurs between two charged objects that are at rest (stationary), is called as Electrostatic Force. Electrostatic Force follows Coulomb's law.
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Examples of Electrostatic Force In Our Daily Life
There are a lot of examples of electrostatic force in our daily life. Some of them are mentioned below.
Commonly Asked Questions
Q1. Find the Electric Force between two protons.
Solution.
$$
F_e=\frac{k_e \cdot e^2}{r^2}
$$
where:
$k_e=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}^2$
$e=1.6 \times 10^{-19} \mathrm{C}$
$r=1 \times 10^{-15} \mathrm{~m}$
1. Square the charge $e$ :
$$
e^2=\left(1.6 \times 10^{-19}\right)^2=2.56 \times 10^{-38} \mathrm{C}^2
$$
2. Square the distance $r$ :
$$
r^2=\left(1 \times 10^{-15}\right)^2=1 \times 10^{-30} \mathrm{~m}^2
$$
3. Substitute these values into Coulomb's Law:
$$
F_e=\frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{1 \times 10^{-30}}
$$
4. Simplify the calculation:
$$
F_e=9 \times 10^9 \times 2.56 \times 10^{-8}
$$
5. Final calculation:
$$
F_e=230.4 \mathrm{~N}
$$
Q2. What is the magnitude of the force that a 25μC charge exerts on a −10μC charge placed at a distance of 8.5cm ?
Solution:
$$
F=k_e \frac{q_1 \cdot q_2}{r^2}
$$
where:
$k_e=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}^2$
$q_1=+25 \mu \mathrm{C}=+25 \times 10^{-6} \mathrm{C}$
$q_2=-10 \mu \mathrm{C}=-10 \times 10^{-6} \mathrm{C}$
$r=8.5 \mathrm{~cm}=0.085 \mathrm{~m}$
1. Substitute the values:
$$
F=9 \times 10^9 \cdot \frac{\left(25 \times 10^{-6}\right) \cdot\left(-10 \times 10^{-6}\right)}{(0.085)^2}
$$
2. Calculate the product of charges:
$$
q_1 \cdot q_2=25 \times 10^{-6} \cdot\left(-10 \times 10^{-6}\right)=-2.5 \times 10^{-10} \mathrm{C}^2
$$
3. Square the distance:
$$
r^2=(0.085)^2=0.007225 \mathrm{~m}^2
$$
4. Substitute these values back:
$$
F=9 \times 10^9 \cdot \frac{-2.5 \times 10^{-10}}{0.007225}
$$
5. Calculate the division:
$$
F=9 \times 10^9 \cdot\left(-3.46 \times 10^{-8}\right)
$$
6. Final multiplication:
$$
F=-311.42 \mathrm{~N}
$$
NOTE: The negative sign in the electric force indicates that the forces between the given two-particle are attractive in nature.
Frequently Asked Questions (FAQs)