Electric field of a half ring: uniformly charged half ring electric field integral.

Описание: 00:00 Given a uniformly charged half ring with charge Q and radius R, we compute the electric field at the center of the half ring using an electric field integral

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00:15 Only the horizontal component of the electric field counts: the total electric field of a half ring must point along the symmetry axis of the ring. This is because the charge distribution is exactly the same on each side of the symmetry axis, so any off-axis contributions to the uniformly charged half ring electric field will cancel.

00:37 Break the ring into point charges: we break the ring into infinitesimal point charges dq that take an infinitesimal arc along the ring. Each charge increment contributes an incremental electric field dE pointing radially away from the charge. We reiterate that the only part of the electric field that matters is the projection of the electric field along the symmetry axis, because every dq above the symmetry axis has a matching dq below the symmetry axis and the y components of their field contributions will cancel. To get the horizontal component of electric field, we tack on the cosine of the angle of elevation for dq.

01:32 How to find linear charge density: linear charge density (lambda) is the charge per unit length in coulombs per meter. We can write the linear charge density of the half ring as lambda=Q/pi*R, since the charge on the half ring is Q and the length is pi*R (half of a circumference). We can also express the total charge on the half ring as Q=lambda*pi*R, or linear charge density times length. This same idea applies to the charge increment itself: dq=lambda*ds, where ds is the small increment of arc for dq. Applying the arc length formula from geometry (s=r*theta), we can say ds=Rd(theta), so the charge increment can be expressed as dq=lambda*Rd(theta).

02:50 Set up the half ring electric field integral: we use the formula for electric field of a point charge and plug in our dq in terms of d(theta). Next, we tack on a factor of cos(theta) to get the x component and express dE_x entirely in terms of theta. We use an integral to sum the electric field contributions, then express the final answer for electric field of a half ring in terms of the total charge on the half ring.