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Two point charges q1 = 2 micro-C and q2 = 1 micro-C are placed at distances b = 1 cm from the y-axis and a = 2 cm from the x-axis respectively (on the respective axes). The electric field vector at point P(a, b) = P(2 cm, 1 cm) makes angle theta with the x-axis such that tan(theta) = K. Find K.
- 1/2
- 2
- 1
- 4
Correct answer: 1
Solution
q1 = 2 micro-C at (0, 1 cm) [on y-axis at b=1 cm]. q2 = 1 micro-C at (2 cm, 0) [on x-axis at a=2 cm]. Point P = (2 cm, 1 cm). E1 (due to q1): displacement from q1(0,1) to P(2,1) = (2,0) cm. Distance r1 = 2 cm = 0.02 m. Magnitude E1 = k*q1/r1² = 9*10⁹ * 2*10⁻⁶ / (0.02)² = 18000/0.0004 = 4.5*10⁷ N/C, directed in +x direction. E2 (due to q2): displacement from q2(2,0) to P(2,1) = (0,1) cm. Distance r2 = 1 cm = 0.01 m. Magnitude E2 = k*q2/r2² = 9*10⁹ * 1*10⁻⁶ / (0.01)² = 9000/0.0001 = 9*10⁷ N/C, directed in +y direction. Total E: Ex = E1 = 4.5*10⁷ N/C, Ey = E2 = 9*10⁷ N/C. tan(theta) = Ey/Ex = 9*10⁷ / 4.5*10⁷ = 2. So K = 2.
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