A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2?

We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values,
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶) / (3 * 10⁻²)²

F = 100 N [ East of positive charge ]

Hope this helps!

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priyankatutor priyankatutor · Answer 2 · 2022-02-01T06:09:06+01:00

The magnitude of the electric force by q1 on q2 is 100 N. Electric forces are directly proportional to the product of charges

What does Coulomb's law state?

The strength of electric forces is directly proportional to the product of charges and inversely proportional to the distance between them.

[tex]F = k \dfrac{q_1q_2}{r^2}[/tex]

Where,

F = electric force

k = Coulomb constant = [tex]\bold {8.99 \times 10^9 \ Nm}[/tex]

q_1, = charge 1 = 5 µC

q_2 = charges = 2 µC

r = distance of separation = [tex]\bold { 3 \times 10^{-2 }\ m}[/tex]

Put the values in the formula,

[tex]F = \bold {8.99 \times 10^9 \ \times } \dfrac{5\times 2}{(\bold { 3 \times 10^{-2 }\ m})^2}\\\\F = 100 \rm \ N[/tex]

Therefore, the magnitude of the electric force by q1 on q2 is 100 N.

Learn more about coulomb's law:

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