\( F_e = kq_1q_2/r^2 \) | \( E = kQ/r^2 \) | \( V = kQ/r \) | \( F = qvB\sin\theta \) | \( F = BIL \) | \( g = GM/r^2 \)
Score: 0 / 6
Q1
Two charges \(q_1 = +4.0\,\mu\text{C}\) and \(q_2 = -6.0\,\mu\text{C}\) are \(0.30\text{ m}\) apart. Calculate the electric force.
Answer:
Solution:
kq₁q₂/r² = 8.99e9 × 4e-6 × 6e-6 / 0.09 = 2.40 N
Q2
What is \(g\) at a distance of \(3R_E\) from Earth's center? (\(g_0 = 9.8\text{ m/s}^2\))
Answer:
Solution:
g = g₀(R/r)² = 9.8(1/3)² = 9.8/9 = 1.09 m/s²
Q3
Electric field lines point:
Q4
A proton (\(q = 1.6\times10^{-19}\text{ C}\)) moves at \(3.0\times10^6\text{ m/s}\) perpendicular to a \(0.50\text{ T}\) field. Calculate the magnetic force (×10⁻¹³ N).
Answer:
Solution:
F = qvB = 1.6e-19 × 3e6 × 0.5 = 2.4e-13 N
Q5
The electric field between parallel plates separated by \(0.02\text{ m}\) with potential difference \(100\text{ V}\) is: