Two point charges Q1 = 2 μ C and Q2 = 1 μ C are placed as shown. The coordinates of the point P are (2 cm, 1 cm). The electric intensity vector at P subtends an angle θ with the positive X-axis. The value of θ is given by

Question

Two point charges Q1 = 2 μ C and Q2 = 1 μ C are placed as shown. The coordinates of the point P are (2 cm, 1 cm). The electric intensity vector at P subtends an angle θ with the positive X-axis. The value of θ is given by (A)  tan θ = 1  (B)  tan θ = 2  (C)  tan θ = 3  (D)  tan θ = 4

Tardigrade Admin · Accepted Answer

Here, Q1 = 2μ C Q2= 1 μ C r1 =AP= 2: cm = 0.02: m r2 = BP = 1 : cm = 0.01 : m From diagram, tan θ = (E2/E1) =(Q2/Q1 ) ×((r1/r2))2 = (1/2) × ((2/1))2 = 2

Q. Two point charges $Q_{1} = 2 μ C$ and $Q_{2} = 1 μ C$ are placed as shown. The coordinates of the point P are (2 cm, 1 cm). The electric intensity vector at P subtends an angle $θ$ with the positive X-axis. The value of $θ$ is given by

$tan θ = 1$

$tan θ = 2$

$tan θ = 3$

$tan θ = 4$

Here, $Q_{1} = 2 μ C$
$Q_{2} = 1 μ C$
$r_{1} = A P = 2 c m = 0.02 m$
$r_{2} = BP = 1 c m = 0.01 m$
From diagram, $tan θ = \frac{E _{2}}{E _{1}}$
$= \frac{Q _{2}}{Q _{1}} \times (\frac{r _{1}}{r _{2}})^{2} = \frac{1}{2} \times (\frac{2}{1})^{2} = 2$

Q. Two point charges Q1​=2μC and Q2​=1μC are placed as shown. The coordinates of the point P are (2 cm, 1 cm). The electric intensity vector at P subtends an angle θ with the positive X-axis. The value of θ is given by

tanθ=1

tanθ=2

tanθ=3

tanθ=4