Question

in fig. 24-28a, what is the potential at point p due to charge q at distance r from p? set v = 0 at infinity. express your answer in terms of given variables, ε₀ and π. in fig. 24-28b, the same charge q has been spread uniformly over a circular arc of radius r and central angle 40°. what is the potential at point p, the center of curvature of the arc? express your answer in terms of given variables, ε₀ and π.

Answer:

First, let's solve the problem for the full circle (Fig. 24 - 28a) and then the arc (Fig. 24 - 28b).

Part 1: Potential at point \( P \) due to charge \( Q \) at distance \( R \) (full circle, Fig. 24 - 28a)

The electric potential at a point due to a point charge is given by \( V=\frac{kq}{r} \), where \( k = \frac{1}{4\pi\epsilon_0} \), \( q \) is the charge, and \( r \) is the distance from the charge to the point. For a full circle, the charge \( Q \) is distributed on the circumference, and the distance from any point on the circle to \( P \) is \( R \) (since \( P \) is at the center of the circle). The potential at \( P \) due to the full circle is the sum of the potentials due to each infinitesimal charge \( dq \) on the circle. Since potential is a scalar quantity, we can simply integrate (or sum) the potentials.

The potential due to an infinitesimal charge \( dq \) at distance \( R \) from \( P \) is \( dV=\frac{1}{4\pi\epsilon_0}\frac{dq}{R} \). Integrating over the entire charge \( Q \) (i.e., summing over all \( dq \)):

\[
V=\int dV=\int_{0}^{Q}\frac{1}{4\pi\epsilon_0}\frac{dq}{R}
\]

Since \( \frac{1}{4\pi\epsilon_0 R} \) is constant with respect to \( dq \), we can factor it out of the integral:

\[
V=\frac{1}{4\pi\epsilon_0 R}\int_{0}^{Q}dq
\]

The integral of \( dq \) from \( 0 \) to \( Q \) is just \( Q \), so:

\[
V = \frac{Q}{4\pi\epsilon_0 R}
\]

Part 2: Potential at point \( P \) due to charge \( Q \) spread over a circular arc (Fig. 24 - 28b)

The key here is that electric potential is a scalar, so we can use the same formula for the potential due to a charge distribution, but we need to consider the fraction of the charge that is on the arc compared to the full circle.

First, let's find the fraction of the circle that the arc represents. The central angle of the arc is \( 40^\circ \). To use this in radians (or to find the fraction of the circumference), we note that a full circle has a central angle of \( 360^\circ \) (or \( 2\pi \) radians). The fraction of the circle (and thus the fraction of the charge, since the charge is uniformly distributed) that is on the arc is \( \frac{\theta}{360^\circ} \), where \( \theta = 40^\circ \).

Let's convert \( 40^\circ \) to radians to make it easier, but we can also work with degrees for the fraction. \( \theta = 40^\circ=\frac{40}{360}\times2\pi=\frac{2\pi}{9} \) radians (but we can also just use the fraction \( \frac{40}{360}=\frac{1}{9} \)).

The charge on the arc is a fraction of the total charge \( Q \). Wait, actually, in the problem, for the arc, the total charge is still \( Q \), but it's spread over the arc. Wait, no—wait, the problem says "the same charge \( Q \) has been spread uniformly over a circular arc". So the total charge on the arc is \( Q \), and we need to find the potential at \( P \) (the center of the arc's curvature) due to this arc.

For a point charge, the potential at distance \( r \) is \( V = \frac{kq}{r} \). For a charge distribution, the potential at a point is the sum (integral) of the potentials due to each infinitesimal charge \( dq \) in the distribution, i.e., \( V=\int\frac{1}{4\pi\epsilon_0}\frac{dq}{r} \), where \( r \) is the distance from \( dq \) to the point \( P \).

In the case of the arc, each infinitesimal charge \( dq \) on the arc is at a distance \( R \) from \( P \) (since \( P \) is the center of the arc, so the radius of the arc is \( R \), and thus the distance from any point on the arc to \( P \) is \( R \)). Therefore, the distance \( r = R \) for all \( dq \) on the arc.

So, similar to the full circle, the potential due to the arc is:

\[
V=\int\frac{1}{4\pi\epsilon_0}\frac{dq}{R}
\]

Since \( \frac{1}{4\pi\epsilon_0 R} \) is constant with respect to \( dq \), we can factor it out:

\[
V=\frac{1}{4\pi\epsilon_0 R}\int dq
\]

But now, \( \int dq \) is the total charge on the arc, which is \( Q \). Wait, that can't be right—wait, no, in the full circle, the charge is \( Q \) on the full circumference, and in the arc, the charge is \( Q \) on the arc. Wait, but the key is that for the potential, since each \( dq \) is at distance \( R \) from \( P \), the potential is just the sum of \( \frac{dq}{4\pi\epsilon_0 R} \) over all \( dq \) on the arc. Since the sum of \( dq \) is \( Q \), the potential should be \( \frac{Q}{4\pi\epsilon_0 R} \)? But that's the same as the full circle? Wait, no, that can't be. Wait, no—wait, no, in the full circle, the charge is on the full circumference, and in the arc, the charge is on the arc, but the distance from each \( dq \) to \( P \) is still \( R \). Since potential is a scalar, it doesn't depend on the direction, only on the magnitude of the distance and the charge. So even if the charge is on an arc, as long as each \( dq \) is at distance \( R \) from \( P \), the potential at \( P \) is the same as if all the charge were at a point at distance \( R \) (but that's only true for potential, not for electric field).

Wait, let's think again. The electric potential at a point due to a charge distribution is given by \( V = \frac{1}{4\pi\epsilon_0}\int\frac{dq}{r} \), where \( r \) is the distance from \( dq \) to the point. In the case of the arc, each \( dq \) is at distance \( R \) from \( P \), so \( r = R \) for all \( dq \). Therefore, the integral becomes \( \frac{1}{4\pi\epsilon_0 R}\int dq \). The integral of \( dq \) over the arc is the total charge \( Q \) on the arc. Therefore, the potential is \( \frac{Q}{4\pi\epsilon_0 R} \). But wait, that's the same as the full circle? But that seems counterintuitive. Wait, no—wait, in the full circle, the charge is on the full circumference, but in the arc, the charge is on the arc, but the distance from each \( dq \) to \( P \) is still \( R \). Since potential is a scalar, it's just the sum of the potentials from each \( dq \), and since each \( dq \) is at the same distance, it's just \( \frac{1}{4\pi\epsilon_0 R} \times \) total charge. So regardless of the shape (full circle or arc), as long as all the charge is at distance \( R \) from \( P \), the potential at \( P \) is \( \frac{Q}{4\pi\epsilon_0 R} \).

Wait, but let's check with the full circle. For a full circle, the electric field at the center is zero (by symmetry), but the potential is not zero. The potential at the center of a charged ring (full circle) with charge \( Q \) and radius \( R \) is indeed \( \frac{Q}{4\pi\epsilon_0 R} \), which matches our calculation. Now, for the arc, which is a portion of the ring, the potential at the center should be the same as the ring's potential if the total charge is the same, because each charge element is at the same distance from the center. The electric field would be different (non - zero for the arc, zero for the full circle), but the potential, being a scalar, just depends on the sum of \( \frac{dq}{r} \), and since \( r = R \) for all \( dq \) in both cases (full circle and arc) when the total charge is \( Q \), the potential is the same.

So, for the arc (Fig. 24 - 28b), the potential at \( P \) (the center of the arc's curvature) is also \( \frac{Q}{4\pi\epsilon_0 R} \)? Wait, but let's confirm with the fraction. Wait, maybe I misread the problem. Wait, the first part: "In Fig. 24 - 28a, what is the potential at point \( P \) due to charge \( Q \) at distance \( R \) from \( P \)? Set \( V = 0 \) at infinity. Express your answer in terms of given variables, \( \epsilon_0 \) and \( \pi \)." Wait, maybe in Fig. 24 - 28a, the charge \( Q \) is a point charge at distance \( R \) from \( P \)? Wait, the figure (a) shows a circle with center \( P \) and radius \( R \), so maybe the charge \( Q \) is distributed on the circle (a ring of charge). Then, for a ring of charge with charge \( Q \) and radius \( R \), the potential at the center is \( \frac{Q}{4\pi\epsilon_0 R} \), which is correct.

Then, for the arc (Fig. 24 - 28b), the charge \( Q \) is spread over a circular arc with radius \( R \) and central angle \( 40^\circ \). Let's find the length of the arc and the length of the full circle. The length of the full circle is \( 2\pi R \), and the length of the arc is \( s = R\theta \), where \( \theta \) is in radians. \( 40^\circ=\frac{40\times\pi}{180}=\frac{2\pi}{9} \) radians. So the length of the arc is \( s = R\times\frac{2\pi}{9} \), and the length of the full circle is \( 2\pi R \). The fraction of the circle (and thus the fraction of the charge, if the charge is uniformly distributed by length) is \( \frac{s}{2\pi R}=\frac{\frac{2\pi}{9}R}{2\pi R}=\frac{1}{9} \). Wait, but the problem says "the same charge \( Q \) has been spread uniformly over a circular arc". So the total charge on the arc is \( Q \), not a fraction. Wait, that must be the case. So the charge per unit length on the arc is \( \lambda=\frac{Q}{s} \), where \( s = R\theta \) ( \( \theta \) in radians). But when calculating the potential at the center, each charge element \( dq=\lambda ds \) is at distance \( R \) from the center, so the potential is \( V=\int\frac{1}{4\pi\epsilon_0}\frac{dq}{R}=\frac{1}{4\pi\epsilon_0 R}\int dq=\frac{Q}{4\pi\epsilon_0 R} \), because \( \int dq = Q \).

Wait, that makes sense. Because potential is a scalar, and each charge element is at the same distance from the center, so the total potential is just the sum of the potentials from each charge element, which is \( \frac{1}{4\pi\epsilon_0 R} \) times the total charge. So regardless of whether the charge is on a full circle or a portion of it (as long as all charge elements are at the same distance from the point where we measure the potential), the potential is the same as if all the charge were a point charge at that distance.

So, for both the full circle (Fig. 24 - 28a) and the arc (Fig. 24 - 28b), the potential at point \( P \) is \( \frac{Q}{4\pi\epsilon_0 R} \).

Final Answer

For the potential at point \( P \) due to charge \( Q \) (either in the full circle or the arc) is \(\boxed{\dfrac{Q}{4\pi\epsilon_0 R}}\)