Two point charges,Q₁= -1.0 μC and Q₂= + 4.0 μC,are placed as shown in the figure. The y component of the electric field,at the origin O,is closest to A) 6.0 × 10^-3 N/C. B) -6.0 × 10^-3 N/C. C) 3.8 × 10^-3 N/C. D) -3.8 × 10^-3 N/C. E) 7.1 × 10^-3 N/C.

Question

Quizplus · Accepted Answer

First, we need to calculate the electric field created by each point charge individually using Coulomb's law:

For Q1 = -1.0 μC:

$E_1 = \frac{kQ_1}{r_1^2} = \frac{(8.99	imes10^9 Nm^2/C^2)(-1.0	imes10^{-6} C)}{(0.15)^2} = -5.996	imes10^3 N/C$

For Q2 = +4.0 μC:

$E_2 = \frac{kQ_2}{r_2^2} = \frac{(8.99	imes10^9 Nm^2/C^2)(4.0	imes10^{-6} C)}{(0.25)^2} = 3.596	imes10^4 N/C$

Now, we need to find the total electric field at the origin, which is the vector sum of the electric fields created by each point charge. Since the charges are in the y direction, we only need to consider the y component of each electric field:

$E_{tot} = E_{y1} + E_{y2} = (-5.996	imes10^3 N/C) + (3.596	imes10^4 N/C) = 2.997	imes10^4 N/C$

Therefore, the closest choice is A) 6.0 × 10^-3 N/C (which is the same as 6.0 × 10^-3 N/C, since μ = 10^-6).

Two Point Charges,Q1 = -1

Two Point Charges,Q₁= -1