Two point charges, Q1 = -2.3 μC and Q2 = +9.3 μC, are placed as shown in the figure. The y component of the electric field, at the origin O, is closest to
In order to find the electric field at the origin O, we need to calculate the electric field contributions from both charges and then combine them using vector addition.
The electric field due to a point charge is given by Coulomb's law: E = k(Q/r^2), where E is the electric field, k is Coulomb's constant (= 8.99 × 10^9 N m^2/C^2), Q is the magnitude of the charge, and r is the distance between the charge and the point where the electric field is being measured.
Let's consider the electric field due to Q1 first:
The distance (r1) from Q1 to the origin O is given in the figure as 2.0 m. We convert it to SI units as follows: r1 = 2.0 m.
The electric field due to Q1 at O is: E1 = k(Q1/r1^2). Substituting the values:
E1 = (8.99 × 10^9 N m^2/C^2) * (-2.3 × 10^(-6) C) / (2.0 m)^2
Next, let's consider the electric field due to Q2:
The distance (r2) from Q2 to the origin O is also given in the figure as 3.0 m. We convert it to SI units as follows: r2 = 3.0 m.
The electric field due to Q2 at O is: E2 = k(Q2/r2^2). Substituting the values:
E2 = (8.99 × 10^9 N m^2/C^2) * (9.3 × 10^(-6) C) / (3.0 m)^2
Now, we need to find the y-components of both electric fields, E1 and E2. Since the charges are aligned along the x-axis, the y-component of the electric field will be zero for E1 (as it only has x-component due to its position) and E2 (as it lies on the x-axis).
Finally, the y-component of the electric field at O is given by the vector sum of E1 and E2, which will be closest to:
E_y = E1_y + E2_y
Since both E1_y and E2_y are zero (as explained earlier), the y-component of the electric field at O is also zero. Therefore, the answer is 0 N/C.