Recitation on Electric Fields Solution
Electricity and Magnetism: PHY-204
Fall Semester 2014
Recitation on Electric Fields Solution 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 4πε0 R2 1 λdl 1 (λ0 cos θ)(Rdθ) = = , 2 4πε0 R 4πε0 R2
dE =
where dl = Rdθ is the arc length and λ is the charge per unit length.
θ
dE
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 1 (λ0 cos θ)(Rdθ) 1 λ0 cos2 θdθ = cos θ = · 4πε0 R2 4πε0 R To obtain the total electric field at P , integrate this expression over the limits −π/2 to π/2, ∫
λ0 Ey = 4πε0 R
π/2
cos2 θdθ −π/2
( ∫ π/2 ) ∫ 1 λ0 1 π/2 = dθ + cos 2θdθ 4πε0 R 2 −π/2 2 −π/2 ( ) π λ0 λ0 = = · 4πε0 R 2 8ε0 R
∵ cos2 θ =
1 + cos 2θ 2
Lets find λ0 , ∫ Q=
∫ dQ =
λdl ∫
π/2
= −π/2
(λ0 cos θ)(Rdθ)
∫
= λ0 R
π/2 −π/2
cos θdθ = 2λ0 R
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 = −Fy ˆj = −qEy ˆj λ0 ˆ = −(3.00µC) j 8ε0 R = −(3.00µC)
10.0µC/m = 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 ⃗ = (3ˆi + 5ˆj) × 105 N/C, the ball uniform electric field as shown in Fig. (2). When E 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 qEx , and ⇒T = sin θ ∑ Fy = qEy + T cos θ − mg = 0.
(1)
(2)
Substituting T in Eq(2), yields ( ) qEx qEy + cos θ − mg = 0 sin θ q(Ex cot θ + Ey ) = mg q =
Date: 8 September, 2014
mg (1.00 × 10−3 )(9.8) = = 1.09 × 10−8 C. Ex cot θ + Ey (3 cot 37◦ + 5) × 105 3
Electricity and Magnetism: PHY-204
Fall Semester 2014
(b) By substituting values in relation for T , derived in (a) qEx sin θ (1.09 × 10−8 )(3 × 105 ) = = 5.44 × 10−3 N. sin 37◦
T =
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
Fall Semester 2014
Recitation on Electric Fields Solution 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 4πε0 R2 1 λdl 1 (λ0 cos θ)(Rdθ) = = , 2 4πε0 R 4πε0 R2
dE =
where dl = Rdθ is the arc length and λ is the charge per unit length.
θ
dE
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 1 (λ0 cos θ)(Rdθ) 1 λ0 cos2 θdθ = cos θ = · 4πε0 R2 4πε0 R To obtain the total electric field at P , integrate this expression over the limits −π/2 to π/2, ∫
λ0 Ey = 4πε0 R
π/2
cos2 θdθ −π/2
( ∫ π/2 ) ∫ 1 λ0 1 π/2 = dθ + cos 2θdθ 4πε0 R 2 −π/2 2 −π/2 ( ) π λ0 λ0 = = · 4πε0 R 2 8ε0 R
∵ cos2 θ =
1 + cos 2θ 2
Lets find λ0 , ∫ Q=
∫ dQ =
λdl ∫
π/2
= −π/2
(λ0 cos θ)(Rdθ)
∫
= λ0 R
π/2 −π/2
cos θdθ = 2λ0 R
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 = −Fy ˆj = −qEy ˆj λ0 ˆ = −(3.00µC) j 8ε0 R = −(3.00µC)
10.0µC/m = 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 ⃗ = (3ˆi + 5ˆj) × 105 N/C, the ball uniform electric field as shown in Fig. (2). When E 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 qEx , and ⇒T = sin θ ∑ Fy = qEy + T cos θ − mg = 0.
(1)
(2)
Substituting T in Eq(2), yields ( ) qEx qEy + cos θ − mg = 0 sin θ q(Ex cot θ + Ey ) = mg q =
Date: 8 September, 2014
mg (1.00 × 10−3 )(9.8) = = 1.09 × 10−8 C. Ex cot θ + Ey (3 cot 37◦ + 5) × 105 3
Electricity and Magnetism: PHY-204
Fall Semester 2014
(b) By substituting values in relation for T , derived in (a) qEx sin θ (1.09 × 10−8 )(3 × 105 ) = = 5.44 × 10−3 N. sin 37◦
T =
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
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