Electric Field Problem Help

[kamehamehaaa]

1)

Two point charges lie on the axis. A charge of 6.4 uC is at the origin, and a charge of -9.5 uC is at x = 10cm . What is the net electric field at (a) c = -4.0 cm and at (b) x = +4.0cm ?


2)

The figure shows a system consisting of three charges q1 = +5.00 uC, q2 = +5.00 uC, q3 = -5.00 uC, and at the vertices of an equilateral triangle of side d = 2.85 cm.

Find the magnitude of the electric field at a point halfway between the charges q1 and q2 .

 

[Kurdt]

What have you tried?

 

[Moetasim]

1) To find the net electric field at a specific point, we can use the principle of superposition, which states that the net electric field at a point is the vector sum of the individual electric fields produced by each charge. In this case, we have two point charges, so we can use the equation E = kq/r^2 to calculate the electric field at each point.

a) At c = -4.0 cm, the distance from the origin to the first charge is 4.0 cm, and the distance from the origin to the second charge is 14 cm. Plugging these values into the equation, we get E1 = (9.0x10^9 Nm^2/C^2)(6.4x10^-6 C)/(0.04 m)^2 = 2.88x10^5 N/C and E2 = (9.0x10^9 Nm^2/C^2)(-9.5x10^-6 C)/(0.14 m)^2 = -2.04x10^5 N/C. To find the net electric field, we take the vector sum of these two fields, E = E1 + E2 = 2.88x10^5 N/C - 2.04x10^5 N/C = 8.4x10^4 N/C.

b) At x = +4.0 cm, the distance from the origin to the first charge is 4.0 cm, and the distance from the origin to the second charge is 6 cm. Plugging these values into the equation, we get E1 = (9.0x10^9 Nm^2/C^2)(6.4x10^-6 C)/(0.04 m)^2 = 2.88x10^5 N/C and E2 = (9.0x10^9 Nm^2/C^2)(-9.5x10^-6 C)/(0.06 m)^2 = -3.17x10^5 N/C. To find the net electric field, we take the vector sum of these two fields, E = E1 + E2 = 2.88x10^5 N/C - 3.17x10^5 N/C = -2.9x10^4 N/C.

2) To find the magnitude of the electric field at a point halfway between