Two charges are placed on the x axis, q1 = 6 uC is placed at
x = 25 cm, and q2 = -10 uC (negative charge) is placed at x =
50 cm. Calc. the electric field due to q1 at the origin point
(0,0). This electric field points left and so is a negative
vector (- i )
To calculate the electric field due to q1 at the origin point (0,0), we can use the principle of superposition. The electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge.
First, we need to calculate the electric field produced by charge q1 at the origin point. The formula to calculate the electric field at a point due to a point charge is given by Coulomb's law:
E = k * q / r^2
where E is the electric field, k is the electrostatic constant (k = 9 x 10^9 Nm^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 calculated.
In this case, the magnitude of charge q1 is 6 uC, and the distance between charge q1 and the origin point is 25 cm = 0.25 m. Substituting these values into the formula, we get:
E1 = (9 x 10^9 Nm^2/C^2) * (6 x 10^-6 C) / (0.25 m)^2
= (9 x 10^9 Nm^2/C^2) * (6 x 10^-6 C) / 0.0625 m^2
= 216 x 10^3 N/C
The electric field due to charge q1 at the origin point is 216 x 10^3 N/C in the leftward direction.
Since the question states that the electric field points left and is a negative vector (-i), we can write the electric field vector as:
E1 = -216 x 10^3 N/C * i
Therefore, the electric field due to q1 at the origin point is -216 x 10^3 N/C in the leftward direction.