Recitation on Electric Fields Solution

Electricity and Magnetism: PHY-204 Fall Semester 2014

1. A line of positive charge is formed into a semicircle of radius R = 60.0 cm as shown in Fig. (1). The charge per unit length along the semicircle is described by the expression

λ = λ0 cos θ. The total charge on the semicircle is 12.0 µC. Calculate the total force on a charge of 3.00 µC placed at the center of curvature P.

FIG. 1

Answer The magnitude of electric ﬁeld at point P due to one segment of bended line of positive charge having a charge dq,

1 dq dE = 2 4πε0 R 1 λdl 1 (λ0 cos θ)(Rdθ) = 2 = 2 , 4πε0 R 4πε0 R where dl = Rdθ is the arc length and λ is the charge per unit length.

Date: 8 September, 2014 1 Electricity and Magnetism: PHY-204 Fall Semester 2014

−→ Figure shows the electric ﬁeld contribution dE at point P due to a single segment of charge at the top of semicircle. Because of the the symmetry of the situation, the

horizontal component of electric ﬁeld dEx = dE sin θ will cancel out. Therefore, only vertical component will contribute

dEy = dE cos θ 2 1 (λ0 cos θ)(Rdθ) 1 λ0 cos θdθ· = 2 cos θ = 4πε0 R 4πε0 R To obtain the total electric ﬁeld at P , integrate this expression over the limits −π/2 to π/2, ∫ π/2 λ0 2 Ey = cos θdθ 4πε0R − ( π/∫2 ∫ ) λ 1 π/2 1 π/2 1 + cos 2θ = 0 dθ + cos 2θdθ ∵ cos2 θ = 4πε R 2 2 2 0 ( )−π/2 −π/2 λ π λ = 0 = 0 · 4πε0R 2 8ε0R

Lets ﬁnd λ0, ∫ ∫ Q = dQ = λdl ∫ π/2 = (λ0 cos θ)(Rdθ) − π/∫2 π/2 = λ0R cos θdθ = 2λ0R −π/2

12µC = 2λ0(60 cm)

∴ λ0 = 10.0µC/m

Therefore, the force on a charge 3 µC placed at the center of curvature P , is −→ F = −Fyˆj = −qEyˆj λ = −(3.00µC) 0 ˆj 8ε0R 10.0µC/m = −(3.00µC) = 0.706(−ˆj). 8(8.85 × 10−12)0.600 m

2. A charged cork ball of mass 1.00 g is suspended on a light string in the presence of a uniform electric ﬁeld as shown in Fig. (2). When E⃗ = (3ˆi + 5ˆj) × 105 N/C, the ball is in equilibrium at θ = 37◦. Find

Date: 8 September, 2014 2 Electricity and Magnetism: PHY-204 Fall Semester 2014

(a) the charge on the ball,

(b) the tension in the string.

FIG. 2

Answer

(a) Free body diagram for a charged ball suspended in the presence of a uniform electric ﬁeld in given below;

Let us sum force components to ﬁnd, ∑ Fx = qEx − T sin θ = 0 (1) qE ⇒ T = x , and ∑ sin θ Fy = qEy + T cos θ − mg = 0. (2)

Substituting T in Eq(2), yields ( ) qE qE + x cos θ − mg = 0 y sin θ

q(Ex cot θ + Ey) = mg mg (1.00 × 10−3)(9.8) × −8 q = = ◦ 5 = 1.09 10 C. Ex cot θ + Ey (3 cot 37 + 5) × 10

Date: 8 September, 2014 3 Electricity and Magnetism: PHY-204 Fall Semester 2014

(b) By substituting values in relation for T , derived in (a) qE T = x sin θ (1.09 × 10−8)(3 × 105) = = 5.44 × 10−3 N. sin 37◦

3. A negatively charged particle −q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in Fig. (3). The particle, conﬁned to move along the x-axis, is moved a small distance x along the axis (where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by ( ) 1 kqQ 1/2 f = . 2π ma3

FIG. 3

Answer First we need to calculate the electric ﬁeld due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring. (a) (b)

−→ Figure (a) shows the electric ﬁeld contribution dE at P due to a single segment of

charge at the top of the ring. This ﬁeld vector can be resolved into components dEx

Date: 8 September, 2014 4 Electricity and Magnetism: PHY-204 Fall Semester 2014

parallel to the axis of the ring and dE⊥ perpendicular to the axis. Figure (b) shows the electric ﬁeld contributions from two segments on opposite sides of the ring. Because of the symmetry of the situation, the perpendicular components of the ﬁeld cancel and only the parallel components will contribute.

dEx = dE cos θ 1 dQ 1 dq = 2 cos θ = 2 2 cos θ. 4πε0 r 4πε0 (x + a )

From the geometry of the ﬁgure, cos θ = x/r = x/(x2 + a2)1/2. So we have

1 xdQ dEx = 2 2 3/2 4πε0 (x + a ) 1 x(λadθ) = 2 2 3/2 4πε0 (x + a ) 1 x(λadθ) = 2 2 3/2 4πε0 (x + a ) 1 xQadθ Q = ∵ λ = 4πε 2πa(x2 + a2)3/2 2πa 0 ∫ 2π ⇒ 1 xQ Ex = 2 2 3/2 dθ 4πε0 2π(x + a ) 0 1 xQ = 4πε 3 x2 3/2 0 a (1 + a2 ) 1 xQ = 3 for x << a. 4πε0 a The magnitude of the force experienced by a charge −q placed along the axis of the ring is

F = −qEx − qxQ = 3 (4πε0a ) − qQ = 3 x. 4πε0a This expression for the force is in the form of Hooke’s law, with an eﬀective spring qQ · constant of k = 3 4πε√0a k Since ω = 2πf = m , we have √ 1 qQ f = 3 . 2π 4πε0ma

Date: 8 September, 2014 5 Electricity and Magnetism: PHY-204 Fall Semester 2014

4. Identical thin rods of length 2a carry equal charges +Q uniformly distributed along their lengths. The rods lie along the x axis with their centers separated by a distance b > 2a (Fig. (4)). Show that the magnitude of the force exerted by the left rod on the right one is ( ) ( ) kQ2 b2 F = ln . 4a2 b2 − 4a2

FIG. 4

Answer Two identical positively charged rods of linear charge density λ having equal lengths 2a lying along x axis. The magnitude of electrostatic repulsion due to one segment of

left rod having a charge dq1, exerted on one segment of right rod having a charge dq2 is, kdq dq dF = 1 2 , r2 where Q dq = λdx = dx , 1 1 2a 1 Q dq = λdx = dx , 2 2 2a 2

and r = x2 − x1 is the separation between two segments as shown in ﬁgure. So we have 2 kQ dx1dx2 · dF = 2 2 4a (x2 − x1)

Date: 8 September, 2014 6 Electricity and Magnetism: PHY-204 Fall Semester 2014

Integrate this expression to obtain the total force ∫ ∫ ∫ ∫ kQ2 dx dx F = dF = 1 2 · 2 − 2 4a x2 x1 (x2 x1) Now ∫ ∫ dx a dx 1 = 1 , − 2 − 2 x1 (x2 x1) −a (x2 x1) let

y = x2 − x1, ⇒ dy = −dx1.

When x1 = −a, y = x2 + a and when x1 = a, y = x2 − a. By these substitutions, integral becomes ∫ − x2−a x2 a −dy 1 = y2 y x2+a x2+a 1 − 1 = − x2 ∫a x2[+ a ] kQ2 1 1 ∴ F = dx − 4a2 2 x − a x + a [∫x2 2 ∫ 2 ] kQ2 b+a dx b+a dx = 2 − 2 4a2 x − a x + a [ b−a 2 b−a 2 ] 2 b+a b+a kQ = ln(x2 − a) − ln(x2 + a) 4a2 [ ( b−a) ( b−a )] kQ2 b +a −a b + a + a = ln − ln 2 − − − 4a [ (b a )a ( b +)]a a kQ2 b b + 2a = ln − ln 2 − 4a [ (b 2a) ( b )] kQ2 b b = ln + ln 2 − 4a ( b 2a ) b + 2a −→ kQ2 b2 F = ln ˆi. 4a2 b2 − 4a2

5. An electric dipole in a uniform horizontal electric ﬁeld is displaced slightly from its equilibrium position as shown in Fig. (5), where θ is small. The separation of the charges is 2a, and each of the two particles has mass m.

(a) Assuming the dipole is released from this position, show that its angular orien- tation exhibits simple harmonic motion with a frequency, √ 1 qE f = · 2π ma

Date: 8 September, 2014 7 Electricity and Magnetism: PHY-204 Fall Semester 2014

(b) Suppose the masses of the two charged particles in the dipole are not the same even though each particle continues to have charge q. Let the masses of the

particles be m1 and m2. Show that the frequency of the oscillation in this case is √ 1 qE(m + m ) f = 1 2 · 2π 2am1m2

FIG. 5

Answer

(a) Torque exerted by electrostatic force on positive charge τ = R×F = −aF sin θ = −qEa sin θ. Torque exerted by electrostatic force on negative charge τ = −qEa sin θ. The electrostatic forces result in a net torque τ = −2qEa sin θ. For small θ, sin ≈ θ, we have τ = −2qEaθ. Also, from analogy of Newton’s second law for rotational motion: d2θ τ = Iα = I , dt2 where I moment of inertia for particles rotating about an axis depends only on their masses and their distances from the axis:

I = ma2 + ma2 = 2ma2 d2θ ⇒ 2ma2 = −2qEaθ dt2 ( ) d2θ qE = − θ, dt2 ma which is the standard equation characterizing simple harmonic motion, with ω2 = qE ma .

Date: 8 September, 2014 8 Electricity and Magnetism: PHY-204 Fall Semester 2014

Then the frequency of oscillation is f = ω/2π, or √ 1 qE f = , 2π ma which is the required result.

(b) When the masses are unequal, the axis of rotation of the dipole passes through the center of mass of the two charges.

We call moment of inertia about this axis ICM, then by using parallel axis theorem:

2 I = ICM + Md

2 ⇒ ICM = I − Md 2 2 − 2 = (m1a + m2a ) (m1 + m2)rcm,

Here rcm is the distance of center of mass from origin which is the point equidis- tance from the two charges on the straight line connecting them.

O q r cm a q a

m1a − m2a (m1 − m2)a rcm = = · m1 + m2 (m1 + m2) then, ( ) − 2 2 (m1 m2)a Icm = (m1 + m2)a − (m1 + m2) (m1 + m2) − 2 2 2 (m1 m2) a = (m1 + m2)a − ( (m1 + m)2) − 2 (m1 m2) 2 = m1 + m2 − a ( (m1 + m2) ) (m + m )2 − (m − m )2 = 1 2 1 2 a2 ( m)1 + m2 4m m = 1 2 a2. m1 + m2

Date: 8 September, 2014 9 Electricity and Magnetism: PHY-204 Fall Semester 2014

Now

d2θ Icm = −2qEaθ ( ) dt2 4m m d2θ 1 2 2 − a 2 = 2qEaθ m1 + m2 dt ( ) 2 ⇒ d θ − qE(m1 + m2) 2 = θ dt 2am1m2 then the frequency of oscillation is √ 1 qE(m + m ) f = 1 2 . 2π 2am1m2

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