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Two charges are placed on the x axis, +5.00 μC at the origin and -13.0 μC at x = 10.0 cm.

1) Find the electric field on the x axis at x = 6.00 cm. Enter a positive value if the electric field points in the +x direction and a negative value if the electric field points in the -x direction.

User Sgt B
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Final answer:

To calculate the electric field at x = 6.00 cm, one must consider the contributions from both the positive and negative charges and use Coulomb's law. The net electric field at this point is found to be -1.65927 x 10^6 N/C, indicating it points in the -x direction.

Step-by-step explanation:

To find the electric field on the x-axis at x = 6.00 cm, we have to consider the contributions from both charges located at the origin (+5.00 μC) and at x = 10.0 cm (-13.0 μC). The electric field due to a point charge is given by the equation E = k|q|/r2, where E is the electric field, k is Coulomb's constant (8.9875 × 109 Nm2/C2), q is the charge, and r is the distance from the charge to the point of interest.

First, calculate the electric field due to the positive charge at the origin:

E+ = (8.9875 × 109 Nm2/C2)(5.00 × 10-6 C) / (0.06 m)2 = 1.24604 × 106 N/C.

This field points in the +x direction.

Next, calculate the electric field due to the negative charge at x = 10.0 cm:

E- = (8.9875 × 109 Nm2/C2)(13.0 × 10-6 C) / (0.04 m)2 = 2.90531 × 106 N/C.

This field points in the -x direction because the charge is negative.

To find the net electric field at x = 6.00 cm, subtract E- from E+ because they are in opposite directions:

Enet = E+ - E- = 1.24604 × 106 N/C - 2.90531 × 106 N/C = -1.65927 × 106 N/C.

The negative sign indicates that the net electric field at x = 6.00 cm points in the -x direction.

User Soulshake
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