Six equal positive charges q sit at the vertices of a regular hexagon with sides oflength R. We remove the bottom charge. The electric field at the center of thehexagon (point P) is:622222 ˆA E j2 ˆB E jˆC E jˆD. E jE E 0
Question
Six equal positive charges q sit at the vertices of a regular hexagon with sides oflength R. We remove the bottom charge. The electric field at the center of thehexagon (point P) is:622222 ˆA E j2 ˆB E jˆC E jˆD. E jE E 0
Solution
The electric field at a point due to a charge is given by Coulomb's law:
E = k*q/r^2
where:
- E is the electric field,
- k is Coulomb's constant,
- q is the charge,
- r is the distance from the charge to the point.
In this case, we have 5 charges at the vertices of a regular hexagon. The electric field at the center of the hexagon due to each charge will be:
E_i = k*q/R^2
where R is the distance from each charge to the center of the hexagon.
Since the hexagon is regular and the charges are equal, the electric fields due to each charge will have the same magnitude but different directions.
If we consider the hexagon to be oriented such that one vertex is at the top, then the electric fields due to the charges at the left and right vertices will cancel each other out, as they are in opposite directions. The same applies to the charges at the top-left and top-right vertices.
The only charge that does not have another charge cancelling its electric field is the one at the top. Therefore, the total electric field at the center of the hexagon will be due to this charge only, and its direction will be upwards.
So, the electric field at the center of the hexagon (point P) is:
E = E_i = k*q/R^2 in the upward direction.
Therefore, the correct answer is B. E = k*q/R^2 ˆj.
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