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in a vacuum two point charges with a magnitude of 1.8nc and -1.8nc are separated by 6mm along x axis. at halfway point of the separation distance , the electric field is

User Guradio
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Final answer:

The net electric field at the midpoint between two point charges of 1.8 nC and -1.8 nC is zero, since their individual electric field contributions cancel each other out.

Step-by-step explanation:

To determine the electric field at the midpoint between two point charges with magnitudes of 1.8 nC and -1.8 nC, we need to calculate the electric field contributions due to each charge separately and then combine them to find the total electric field at that point. Since the charges are equal and opposite, and the point of interest is equidistant from both charges, their individual electric fields at the midpoint will be equal in magnitude but opposite in direction.

For a point charge, the electric field (E) at a distance (r) is given by Coulomb's law:

E = k * |q| / r2

where k is Coulomb's constant (k = 8.99 x 109 Nm2/C2) and q is the charge. In this scenario, r is half the separation distance, which is 3 mm or 3 x 10-3 m. Substituting the charge magnitude and the distance into the equation, we get:

E = (8.99 x 109 Nm2/C2) * (1.8 x 10-9 C) / (3 x 10-3 m)2

The electric field due to each charge will be:

E = 5.4 x 105 N/C

Because the two electric fields are equal in magnitude and opposite in direction at the midpoint, they will cancel each other out, resulting in a net electric field of zero at the halfway point.

User Matsev
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