Question

Three −9mC point charges are located at (0,0),(3 m,3 m), and (3 m,−3 m). What is the magnitude of the electric field at (3 m,0) ? a) 0.9×10 7

N/C e) 3.6×10 7
N/C b) 1.2×10 7
N/C f) 5.4×10 7
N/C c) 1.8×10 7
N/C g) 10.8×10 7
N/C d) 24×10 7
N/C

Answer 1

The magnitude of the electric field at (3 m,0) is 6 * 10^7 N/C.

Step-by-step explanation:

To calculate the magnitude of the electric field at (3 m,0), we need to consider the electric fields created by each of the three point charges. The electric field created by a point charge can be found using the formula:

E = krac{Q}{r^2}

where E is the electric field, k is the Coulomb's constant (8.99 × 10^9 Nm^2/C^2), Q is the charge, and r is the distance between the charge and the point at which we want to find the electric field.

Using this formula, we can calculate the electric fields created by the three-point charges and find their vector sum to get the overall electric field at (3 m,0).

First, let's find the electric field created by the charge at (0,0). Since it is located at the origin, the distance is 3 m. The electric field created by this charge is:

E1 = (8.99 × 10^9 Nm^2/C^2) * rac{9mC}{(3m)^2} = 3 * 10^7 N/C

Next, let's find the electric field created by the charge at (3 m,3 m). Using the distance formula, the distance between this charge and (3 m,0) is also 3 m. The electric field created by this charge is:

E2 = (8.99 × 10^9 Nm^2/C^2) * rac{-9mC}{(3m)^2} = -3 * 10^7 N/C

Finally, let's find the electric field created by the charge at (3 m,-3 m). Again, the distance between this charge and (3 m,0) is 3 m. The electric field created by this charge is:

E3 = (8.99 × 10^9 Nm^2/C^2) * rac{-9mC}{(3m)^2} = -3 * 10^7 N/C

To find the overall electric field at (3 m,0), we need to add these three electric fields together. Since the vector sum of -3 * 10^7 N/C, -3 * 10^7 N/C, and 0 N/C gives us -6 * 10^7 N/C, the magnitude of the electric field at (3 m,0) is 6 * 10^7 N/C.