A point charge of 4 μC is placed at the origin. There is an external uniform field E = 400 i N/C. What is the net force on a 5 μC charge placed at (5 m, 3 m)
??? i + ??? j N
To find the net force on the 5 μC charge, we can use Coulomb's law which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for the electric force between two charges q1 and q2 is given by:
F = k * |q1 * q2| / r²
where:
- F is the force between the charges,
- k is the electrostatic constant (k ≈ 9 × 10^9 N m²/C²),
- |q1 * q2| is the product of the magnitudes of the charges,
- r is the distance between the charges.
In this case, the point charge at the origin has a charge of 4 μC, and the charge at (5 m, 3 m) has a charge of 5 μC.
The distance between the two charges can be calculated using the Pythagorean theorem:
r = √((x₂ - x₁)² + (y₂ - y₁)²)
where:
- x₁ and y₁ are the coordinates of the first point (in this case, the origin),
- x₂ and y₂ are the coordinates of the second point (in this case, (5 m, 3 m)).
Substituting the given values, we have:
x₁ = 0 m, y₁ = 0 m, x₂ = 5 m, y₂ = 3 m.
Now we can calculate the distance:
r = √((5 - 0)² + (3 - 0)²) = √(5² + 3²) = √34 ≈ 5.83 m.
Substituting the values into Coulomb's law, we have:
F = (9 × 10^9 N m²/C²) * |4 μC * 5 μC| / (5.83 m)²
Calculating the product of the charges:
|4 μC * 5 μC| = |20 × 10^-6 C²| = 20 × 10^-12 C².
Now substituting the values into the formula:
F = (9 × 10^9 N m²/C²) * (20 × 10^-12 C²) / (5.83 m)²
Simplifying the equation:
F ≈ 2.41 × 10^-3 N
Therefore, the net force on the 5 μC charge is approximately 2.41 × 10^-3 N in the x-direction (i) and 0 N in the y-direction (j).