To determine the electric field at point C, we need to first calculate the individual electric fields at point C due to each charge at points A and B, and then add them up vectorially.
The electric field due to the +4nC charge at point A can be calculated using Coulomb's law:
E1 = k(q1/r1^2) = (9 x 10^9 N m^2/C^2)(4 x 10^-9 C)/(100 cm)^2 = 36 N/C
where k is the Coulomb constant, q1 is the charge of the +4nC charge, and r1 is the distance from point A to point C.
Similarly, the electric field due to the -6nC charge at point B can be calculated as:
E2 = k(q2/r2^2) = (9 x 10^9 N m^2/C^2)(-6 x 10^-9 C)/(100 cm)^2 = -54 N/C
where q2 is the charge of the -6nC charge, and r2 is the distance from point B to point C.
Note that the negative sign in E2 indicates that the direction of the electric field due to the negative charge at point B is opposite to the direction of the electric field due to the positive charge at point A.
To add the two electric fields together, we can use vector addition. Since the electric fields at points A and B are pointing in opposite directions, we can subtract them:
E = E1 - E2 = 36 N/C - (-54 N/C) = 90 N/C
Therefore, the electric field at point C due to the +4nC charge at point A and the -6nC charge at point B is 90 N/C, pointing towards point A (since the electric field due to the positive charge is stronger than the electric field due to the negative charge, the overall direction of the field is towards point A).
Suppose you have two charges, +4nC at point A and -6nC at point B that are separated by a distance of 100cm. What would be the electric field at point C that is 100cm away from point B?
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