Physics (2): Problem set 1 solutions

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1 Physics (2): Problem set solutions PHYS 04 Problem : Two identical charges q = nc are located on the x-axis at positions 2 cm and 2 cm. What is the electric field at the origin (centre between the two point charges)? What is the electric field at position x = 4 cm? What is the electric force (magnitude and direction) on a third charge q placed on the y-axis a distance 2 cm from the origin? Solution: The electric field at the centre between the two identical charges is just zero (by symmetry). At position x = 4 cm, the electric field is the sum of two components: E x = k q, E d 2 2x = k q d 2 2 where d = 0.02 m and d 2 = 0.06 m are the distances to the point charges, and both fields are in the same direction (positive x-axis). Thus E = N/C. The electric force on the negative charge placed on the y-axis a distance 2 cm from the origin has only a y-component by symmetry (i.e. F x = 0). The y-component is F y = 2 k q2 d 2 sin45 = N where d is the distance between the negative charge and the other two positive charges. The negative sign indicates the direction of the force is along the negative y-axis. y F F F 2 2 cm 45 x 2 cm 2 cm Figure : problem Problem 2: A positive charge q is placed exactly in the middle between two other identical charges q. Show that the particle in the middle exhibits a simple harmonic motion if it is displaced from its equilibrium position a distance ǫ d, where d is the distance between the middle charge and the other two charges, and find its period. What would happen if the charge was negatively charged instead?

2 dǫ dǫ net force ǫ Figure 2: problem 2 Solution A positively charged particle in the middle between two other positively charged particles is displaced a distance ǫ on the positive x-direction, we have the total force on it is: F = = kq 2 (dǫ) k q 2 2 (dǫ) 2 kq 2 d 2 (ǫ/d) kq 2 2 d 2 (ǫ/d) 2 = kq2 d 2 [ (ǫ/d) 2 (ǫ/d) 2] = kq2 d 2 [ (2ǫ/d)(2ǫ/d)O(ǫ 2 /d 2 ) ] where in the last equality we ignored O(ǫ 2 /d 2 ) terms as they are much smaller than O(ǫ/d) terms. We used in this expansion the fact that (±x) n = ±nx for small x. Thus F = kq2 d 2 [ 2ǫ/d2ǫ/dO(ǫ 2 /d 2 ) ] = kq2 d 2 [4ǫ/d] = 4kq2 d 3 ǫ = Cǫ where the constant C = 4kq 2 /d 3. This equation looks like Hooks law F = Cx. Hence the particle will exhibit a simple harmonic motion with angular frequency ω = C/m. The period of oscillation is T = 2π/ω. The middle particle is said to be in a stable equilibrium. Note: If the charge in the middle is negative, then it would be in an stable equilibrium. In fact it would exponentially drift to the charge to which it is initially closest (like a ball on the top of a hill). Problem 3: Find the expression for the electric field a distance r from a long wire carrying a linear charge distribution λ. 2

3 y δe y δ E θ δe x δq = λδx x 2 r 2 r θ λ x δx x Solution Figure 3: problem 3 We divide the wire into small elements of length dx of charge dq = λdx. By symmetry we can deduce thattheelectric fieldhasonlyay-component(seefigure3)since thecomponent de x cancels against the x-component of the field from the charge at x. Hence we only need to compute the y-component, which is given by: de y = desinθ = kdq x 2 r 2 r x2 r = krλ dx 2 (x 2 r 2 ) 3/2 The total electric field can be found by integrating the above expression over all values of x. If the wire has length 2L, we have: L dx E y = krλ L (x 2 r 2 ) 3/2 This integral can be performed by integration. You can check by differentiation that: dx (x 2 r 2 ) = x 3/2 r 2 x 2 r 2 Thus we find: [ x E y = krλ r 2 x 2 r 2 λ 2L = r L2 r 2 λ = 2πǫ 0 r r2 /L 2 We are measuring the field in a point belonging to the axis of the wire. We shall set L at the end of the calculation for a very long wire. 3 ] L L

4 If we are measuring the electric field at a point very close the finite wire (r L), or equivalently if the wire is infinitely long, we simplify the result to (r L): E = λ 2πǫ 0 r which is exactly the expression we obtained using Gauss s law. Problem 4: A positive charge q is placed in the cavity within a neutral hallow conducting sphere of inner radius R and outer radius R 2 (R < R 2 ).. Draw a diagram representation for the distribution of charges in the sphere. 2. Calculate the magnitude electric field in all regions of the space. 3. What is the total charge on the inner surface of the conductor, and on the outer surface? 4. Draw electric field lines. 5. Calculate the electric potential at every point in space, assuming that the potential reference is infinity. The sphere is connected to the ground. Redraw the distribution of charges and electric field lines. How would the potential differ in this case? Solution. The drawing is shown in figure To compute the electric field we use Gauss s law. For this we choose a sphere of radius r which is concentric with the hallow sphere and charge q. By symmetry the electric field (if any) is constant over this surface and is perpendicular to it, hence the flux over Gauss s surface is just: Φ E = E.dA = EdA = E da = EA = E4πr 2 Then by Gauss s law we have: Thus: Therefore: We distinguish the following cases: Φ E = q enc ǫ 0 E4πr 2 = q enc ǫ 0 E = q enc r 2 0 < r < R : This is the region within the cavity of the conductor. Here the charge enclosed is just q, which means: E = q r 2 4

5 Figure 4: problem 4 R < r < R 2 : This is the region within the conductor. Immediately we notice that E = 0 (inside a conductor). This automatically means that q enc = 0, which implies that the charge that accumulates in the inner surface of the conductor must cancel out the charge q. Thus the charge in the inner surface is q. r > R 2 This is the region outside the sphere. Since the conductor is neutral the charge on the outer surface must be q. and thus the total charge enclosed by Gauss s surface in this case is just q enc = q, and we obtain: E = q r 2 The electric field as a function of r is shown in figure 5 3. The total charge on the inner surface of the conductor is just q, and that on the outer surface is just q. Note that even if the shape of the conductor is irregular this property would still be correct (i.e. charge in the inner surface exactly equals the opposite of the charge in the outer surface, which equals the charge in the cavity). 4. Field lines are drawn in fig The electric potential V is given by V = Edr. We shall fix the potential to be zero at infinity. We discuss the following cases: 5

6 E(r) R R 2 r Figure 5: E as a function of r. r > R 2 : We call this region A. Here integrating the expression for E we obtain ( dr/r 2 = /r) V A = q r C where the constant of integration is chosen so as to set V = 0 when r. Clearly C = 0. We shall need the expression for the voltage at the boundary with the outer surface: V A (R 2 ) = q R 2 R < r < R 2 : Here the electric field is just zero, which means the potential is constant (the constant of integration): V B = C 2. The constant C 2 is fixed by the requirement of the continuity of the electric potential across the boundaries of the regions. Particulary when r = R 2 we must have V A (R 2 ) = V B (R 2 ), which gives: which is constant. V B (r) = q R 2 0 < r < R : We call this region C, and here we again have: V C = q r C 3 and once again the constant C 3 is fixed by the continuity of the potential across the boundary at r = R. Here: V C (R ) = V B (R ), which means: Isolating C 3 and substituting we find that: q R C 3 = q R 2 V C (r) = q 6 [ r R 2 R ]

7 V(r) V 0 R R 2 r Figure 6: The potential V(r) as a function of r. In the picture V 0 = q R 2 If the sphere is grounded then the positive charges on the outer surface are grounded and we obtain the distribution and field lines shown in figure 7. Figure 7: The sphere grounded. The electric field in this case is zero everywhere except for the region 0 < r < R, where it has the same value as before. Hence the potential is constant for r > R and is just equal to zero. However inside the cavity within the conductor it simply becomes: V = q r C 4 where the constant C 4 is chosen so as the potential at the boundary at r = R is a continuous 7

8 function. Thus we find: for r < R and zero otherwise. V = q ( r ) R 8