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Coulomb's constant, denoted as \( k \), is a fundamental value in electrostatics. It appears in Coulomb's law, which describes the force between two point charges. The value of Coulomb's constant is \( 8.99 \times 10^9 \mathrm{~N \ m^2/ C^2} \). This constant is crucial for calculating the electric field created by a charge.
Coulomb's constant can be understood as representing the electric force per unit charge squared. It tells us how strong the force between charges is based on their separation distance.
By using this constant, we can easily determine the magnitude of the electric field (E) around a point charge (q) at a specific distance (r). Thus, Coulomb's constant is a key part of the formula for the electric field: \( E = \frac{kq}{r^2} \).
Learning this constant is beneficial for understanding other concepts such as electric potential and forces between multiple charges. Applying Coulomb's constant correctly helps with solving problems in electrical and magnetism fields.

The electric field is a vector field that represents the force per unit charge exerted on a small positive test charge at any point in space. It is created by electric charges and can act at a distance.
Mathematically, for a point charge, the electric field (E) is expressed as: \( E = \frac{kq}{r^2} \), where:

• E: Electric field
• k: Coulomb's constant
• q: Charge
• r: Distance from the charge

The direction of the electric field is radial. It points away from a positive charge and towards a negative charge.
In our original problem, the electric field at 50 cm away from a charge is given as 2.0 N/C. Understanding the relationship between the electric field, the charge, and the distance helps us determine how charges interact and affect their surroundings.
The electric field is not just a theoretical concept. It's vital for analyzing circuits, understanding how electricity flows, and even engineering applications like designing electronic devices.

To find the amount of charge (q) in a physics problem, we often start with the electric field formula: \( E = \frac{kq}{r^2} \). Given the electric field and the distance, we can rearrange the formula to solve for q: \( q = \frac{E r^2}{k} \).
Let's apply this to our problem step-by-step:

• We know the electric field (E) is 2.0 N/C and the distance (r) is 50 cm, which is 0.50 meters.
• We substitute the values into the rearranged formula: \( q = \frac{(2.0 \mathrm{~N/C}) (0.50 \mathrm{~m})^2}{8.99 \times 10^9 \mathrm{~N m^2/ C^2}} \).
• Calculating this gives us: \( q = \frac{2.0 \times 0.25}{8.99 \times 10^9} \).
• This results in \( q \approx 5.56 \times 10^{-11} \mathrm{~C} \).

Thus, the amount of charge is approximately \( 5.56 \times 10^{-11} \mathrm{~C} \). Practicing these calculations helps solidify your understanding of electric fields and charge. It also prepares you for more complex scenarios where multiple charges and forces are involved. Remember, correctly substituting values and following algebraic steps ensures precise and meaningful results.