Two charges are placed on the x axis. One of the charges (q1 = +7.91C) is at x1 = +3.00 cm and the other (q2 = -27.3C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.
find the E from each, and add. Remember the direction of E fields (the direction a positive test charge will go, which is away from positive charges, and toward negative charges). Sketch the diagram, to make certain you add them correctly (they should be in the same direction inthis problem) for b, and in opposite directions in a.
To find the net electric field at a particular point due to two charges, we can use the principle of superposition. The electric field due to each charge is calculated separately, and then their contributions are added vectorially.
The electric field due to a point charge at a distance r from the 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 * 10^9 Nm^2/C^2),
q is the charge, and
r is the distance.
Let's calculate the electric field due to each charge at the given positions:
(a) At x = 0 cm:
For q1, the distance r1 from the charge to the point is 3.00 cm = 0.03 m.
For q2, the distance r2 from the charge to the point is 9.00 cm = 0.09 m.
So, the electric field due to q1 at x = 0 cm is:
E1 = (k * q1) / r1^2
And the electric field due to q2 at x = 0 cm is:
E2 = (k * q2) / r2^2
To find the net electric field at x = 0 cm, we need to add the contributions from both charges:
E_net = E1 + E2
(b) At x = 6.00 cm:
Now the distance r1 from q1 to the point is 6.00 cm = 0.06 m.
And the distance r2 from q2 to the point is also 6.00 cm = 0.06 m.
So, the electric field due to q1 at x = 6.00 cm is:
E1 = (k * q1) / r1^2
And the electric field due to q2 at x = 6.00 cm is:
E2 = (k * q2) / r2^2
To find the net electric field at x = 6.00 cm, we again need to add the contributions from both charges:
E_net = E1 + E2
Now let's calculate the values:
(a) At x = 0 cm:
E1 = (9 * 10^9 Nm^2/C^2 * 7.91C) / (0.03m)^2
E2 = (9 * 10^9 Nm^2/C^2 * -27.3C) / (0.09m)^2
E_net = E1 + E2
(b) At x = 6.00 cm:
E1 = (9 * 10^9 Nm^2/C^2 * 7.91C) / (0.06m)^2
E2 = (9 * 10^9 Nm^2/C^2 * -27.3C) / (0.06m)^2
E_net = E1 + E2
By calculating these values, you will determine the magnitude and direction (positive or negative sign) of the net electric field at x = 0 cm and x = 6.00 cm.