A thin glass rod is bent into a semicircle of radius R = 12.9 cm. A charge Q = 4.95 nC is uniformly distributed along the upper half, and a charge −Q is uniformly distributed along the lower half as shown in the figure. Find the y-component of the electric field at point P, the center of the semicircle. (You may enter your calculation using scientific notation. Include a minus sign if necessary.) N/C
Question
A thin glass rod is bent into a semicircle of radius R = 12.9 cm. A charge Q = 4.95 nC is uniformly distributed along the upper half, and a charge −Q is uniformly distributed along the lower half as shown in the figure. Find the y-component of the electric field at point P, the center of the semicircle. (You may enter your calculation using scientific notation. Include a minus sign if necessary.) N/C
Solution
To solve this problem, we need to use the formula for the electric field due to a line of charge, which is given by:
E = k * λ / r
where:
- E is the electric field,
- k is Coulomb's constant (9 * 10^9 N m^2/C^2),
- λ is the linear charge density (charge per unit length), and
- r is the distance from the line of charge.
In this case, the line of charge is the semicircle, and the distance r is the radius R of the semicircle.
First, we need to calculate the linear charge density λ. Since the charge Q is uniformly distributed along the semicircle, the linear charge density is given by:
λ = Q / (πR)
Next, we substitute the values given in the problem into the formula for the electric field:
E = k * λ / R = (9 * 10^9 N m^2/C^2) * (4.95 * 10^-9 C / (π * 0.129 m)) / 0.129 m
Solving this gives:
E = 0 N/C
The reason the electric field is zero is because the positive and negative charges on the semicircle create electric fields that are equal in magnitude but opposite in direction. At the center of the semicircle, these electric fields cancel each other out, resulting in a net electric field of zero.
Therefore, the y-component of the electric field at point P is also zero.
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