Question

Two charges are placed on the x axis. One of the charges (q1 = +6.27μC) is a x = +3.00 cm and the other (q2 = -20.4 μC) 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) +6.00 cm.

I am understanding part a but I am lost on part b.

Answer 1

direction of the net electric field will depend on the signs and locations of the charges. If the net electric field is positive, it means the field points away from the charges. If the net electric field is negative, it means the field points towards the charge

Step-by-step explanation:

E = k * q / r^2

where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point where we want to find the electric field.

Let's calculate the net electric field at the given points:

(a) x = 0 cm:

For the first charge, q1 = +6.27 μC, the distance r1 = 3.00 cm = 0.03 m.

The electric field due to q1 at x = 0 cm is:

E1 = k * q1 / r1^2

For the second charge, q2 = -20.4 μC, the distance r2 = 9.00 cm = 0.09 m.

The electric field due to q2 at x = 0 cm is:

E2 = k * q2 / r2^2

To find the net electric field, we need to add the electric fields vectorially:

E_net = E1 + E2

(b) x = +6.00 cm:

For the first charge, q1 = +6.27 μC, the distance r1 = 6.00 cm = 0.06 m.

The electric field due to q1 at x = +6.00 cm is:

E1 = k * q1 / r1^2

For the second charge, q2 = -20.4 μC, the distance r2 = 6.00 cm = 0.06 m.

The electric field due to q2 at x = +6.00 cm is:

E2 = k * q2 / r2^2

To find the net electric field, we need to add the electric fields vectorially:

E_net = E1 + E2