Find Electric Field Magnitude at Center of Charged Semi-Circle

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In summary: Also, don't forget the integral sign in front of your expression.In summary, the task is to find the magnitude of the electric field at point P due to a semicircle of uniformly distributed positive charge Q with radius a. Using charge density per unit length, the electric field strength at P can be expressed as sin(theta)kQ/(a^3 * pi), where theta is the angular position along the semicircle. By integrating this expression from theta = 0 to pi, the final answer is 2kQ/(a^3 * pi). However, the initial calculation is incorrect as Q represents the total charge of the semicircle and the charge density per unit length needs to be used instead.
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cp255
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Homework Statement


Positive charge Q is uniformly distributed around a semicircle of radius a.
attachment.php?attachmentid=65621&stc=1&d=1389750848.jpg


Find the magnitude of the electric field at the center of curvature P.
Express your answer in terms of the given quantities and appropriate constants.

Homework Equations





The Attempt at a Solution



So here is my thinking. The charge per unit length of the circle would be Q/(a*pi) because a*pi is the arc length of the semi-circle. Next I assumed that the x-component is zero due to symmetry so I only need to calculate the force in the y direction. Therefore the magnitude of the electric field created by each point on the circle at the origin would be kQ/(a^2 * pi * a). The y-component of that electric field would be sin(theta)kQ/(a^3 * pi). I integrated this from theta = 0 to pi and the answer was 2kQ/(a^3 * pi). Where did I go wrong?
 

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cp255 said:

Homework Statement


Positive charge Q is uniformly distributed around a semicircle of radius a.
attachment.php?attachmentid=65621&stc=1&d=1389750848.jpg


Find the magnitude of the electric field at the center of curvature P.
Express your answer in terms of the given quantities and appropriate constants.

Homework Equations


The Attempt at a Solution



So here is my thinking. The charge per unit length of the circle would be Q/(a*pi) because a*pi is the arc length of the semi-circle. Next I assumed that the x-component is zero due to symmetry so I only need to calculate the force in the y direction.

Right up to this point. Except I would replace "force" with "component of electric field strength". Field strength is not the same thing as force.

Therefore the magnitude of the electric field created by each point on the circle at the origin would be kQ/(a^2 * pi * a). The y-component of that electric field would be sin(theta)kQ/(a^3 * pi). I integrated this from theta = 0 to pi and the answer was 2kQ/(a^3 * pi). Where did I go wrong?

This part is wrong. "Q" here represents total charge of the semicircle. You need to find a way to represent the charge of a small element along the ring, and then work out the electric field due to it at P.

Represent the charge density per unit length by ##\sigma##. Now what is the electric field strength exerted by a charged element of length ##dl## at point P? What is the y-component? Using radian measure, how is the arc length ##l## related to ##a## and ##\theta##? Before integrating you need to make sure the only variable is ##\theta##.
 

1. What is the formula for calculating the electric field magnitude at the center of a charged semi-circle?

The formula for calculating the electric field magnitude at the center of a charged semi-circle is E = kQ/R, where k is the Coulomb's constant, Q is the charge of the semi-circle, and R is the radius of the semi-circle.

2. How do you determine the direction of the electric field at the center of a charged semi-circle?

The direction of the electric field at the center of a charged semi-circle is determined by the direction of the electric field due to each individual point charge that makes up the semi-circle. The direction of the electric field due to a point charge is always away from a positive charge and towards a negative charge.

3. Can the electric field at the center of a charged semi-circle be zero?

Yes, the electric field at the center of a charged semi-circle can be zero if the total charge of the semi-circle is zero or if the distance from the center of the semi-circle to the individual point charges is equal on both sides, resulting in the electric fields cancelling each other out.

4. How does the electric field magnitude at the center of a charged semi-circle change with an increase in the charge or radius?

The electric field magnitude at the center of a charged semi-circle is directly proportional to both the charge and the radius. This means that an increase in either the charge or the radius will result in an increase in the electric field magnitude at the center of the semi-circle.

5. Can the electric field at the center of a charged semi-circle be negative?

Yes, the electric field at the center of a charged semi-circle can be negative if the charge of the semi-circle is negative or if the direction of the electric field due to each individual point charge is towards the center of the semi-circle. In this case, the negative sign indicates the direction of the electric field, not the magnitude.

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