Question

A uniformly charged insulating rod is bent into the shape of a semicircle of radius R = 5 cm. If the rod has a total charge of Q = 3.10-9C, find the magnitude and direction of the electric field at O, the center of the circle.

Answer 1

Hi there!

We can begin by using Coulomb's Law:

`E = (kq)/(r^2)`

k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)

E = Electric field strength (N/C)
r = distance from point (m)

q = charge (C)

Since this is a continuous charge, we must use calculus.

We can express this as the following:

`q = \lambda L`

λ = Linear charge density (C/m)

L = Length of rod (m)

Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:

`dq = \lambda ds\\\\dq = \lambda rd\theta`

Our new equation is:

`dE = (kdq)/(r^2)\\\\dE = (k\lambda rd\theta)/(r^2)`

However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:

`dE = (k\lambda rd\theta)/(r^2)cos\theta\\\\dE = (k\lambda)/(r)cos\theta d\theta`

Integrate. For a semicircle, the bounds will be from -π/2 to π/2.

`E = (k\lambda)/(r)\int\limits^{(\pi)/(2)}_{-(\pi)/(2)}} {cos\theta} \, d\theta\\\\E = (k\lambda)/(r)sin\theta\left \|{(\pi)/(2)} \atop {-(\pi)/(2)}} \right. \\\\E = (k\lambda)/(r)(1 - (-1)) = (2k\lambda)/(r)`

We need to solve for λ, which is Q/ L:

`\lambda = (3.10 * 10^(-9) C)/(\pi (0.05)) = 1.9735 * 10^(-8) (C)/(m)`

Now, plug and solve for the electric field strength:

`E = (2(8.99* 10^9)(1.9735* 10^(-8)))/(0.05) = \boxed{7096.783 (N)/(C)}`

**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.