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PHYSICS 120 : ELECTRICITY AND MAGNETISM TUTORIAL QUESTIONS GAUSS’ LAW Question 43 A thin-walled metal sphere has a radius of 25.0 cm and carries a charge of 2.00 × 10−7 C. Find E for a point (a) inside the sphere, (b) just outside the sphere, and (c) 3.00 m from the centre of the sphere. (a) Inside the shell there is no charge, so ψ= QENC =0 ǫ0 hence the field inside the shell is zero. (b) To find the field just outside the shell, choose the Gaussian surface to be a sphere, with surface area 4πr2 , so using Gauss’ Law: QENC ψ= = E × (4πr2 ) ǫ0 1 QENC ∴E = 4πǫ0 r2 2 × 10−7 = (9 × 109 ) 6.25 × 10−2 = 2.88 × 104 N C−1 (c) At a distance r = 3.00 m away: 1 QENC 4πǫ0 r2 2 × 10−7 = (9 × 109 ) 3.02 −1 = 200 N C E = Question 44 Two charged concentric spheres have radii of 10.0 cm and 15.0 cm. The charge on the inner sphere is 4.00×10−8 C and that on the outer sphere −2.00×10−8 C. Find E at (a) r = 12.0 cm and (b) r = 20.0 cm. page 1 of 5 PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS (a) At a distance r = 12 cm, the field due to the outer sphere is zero, hence the enclosed charge is the charge of the inner sphere: QENC = QI . The electric field is then QENC ǫ0 A 1 4 × 10−8 = 4πǫ0 (12 × 10−2 )2 = 2.50 × 104 N C−1 E = (b) Let EI and EO be the fields due to the inner and outer spheres. The fields are EI = 1 QI 4πǫ0 r2 4 × 10−8 (20 × 10−2 )2 = 9.00 × 103 N C−1 outwards −2 × 10−8 = (9 × 109 ) (20 × 10−2 )2 = −4.50 × 103 N C−1 inwards = (9 × 109 ) EO By the principle of superposition, the resultant field is E = EI + EO = 4.50 × 103 N C−1 radially outwards page 2 of 5 PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS Question 45 A small sphere whose mass m is 1 mg carries a charge q of 0.02 µC. It hangs from a silk thread which makes an angle of 30° with a large, charged conducting sheet as shown. Calculate the charge density σ for the sheet. Since the system is in equilibrium, the forces are balanced as so Ty = W and Tx = F , where T is the tension in the string, W is the weight of the sphere and F is the repulsive force between the sphere and the sheet. Since the electric field due to the sheet is E= σ 2ǫ0 whereσ is the surface charge density. The force F is then given by E= σ F = q 2ǫ0 The x- and y-components of the tension are 1 Tx = T sin 30° = T 2√ 3T Ty = T cos 30° = 2 √ 1 3T and F = T ∴W = 2 2 2 ⇒ 2F = √ W 3 1 F = √ mg 3 The surface charge density σ is then 2ǫ0 1 √ mg q 3 (2) × (8.85 × 10−12 ) × (1 × 10−6 ) × (9.8) √ = 3 × (0.02 × 10−6 ) = 5.01 × 10−9 C m−2 σ = page 3 of 5 PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS Question 46 Two uniformly charged planes with surface charge densities of 3.00×10−9 C m−2 and −1.00× 10−9 C m−2 lie parallel to each other a distance 8.00 cm apart in a vacuum. Draw the field lines between the plates and behind each plate and find E everywhere. (Hint: use superposition.) Assume infinite sheets (so, therefore, no edge effects) and use the result E = σ/(2ǫ0 ) for a plane surface with surface charge density σ. For E1 , the electric field depends on the electric field EA due to sheet A and the electric field EB due to B. The magnitude of E1 is |E1 | = ||EA | − |EB || 1 = ||σA | − |σB || 2ǫ0 1 = × ((3 − 1) × 10−9 ) −12 8.85 × 10 = 113 N C−1 2π 1 ||σA | − |σB || in terms of the Coulomb constant ke = 2π 2ǫ0 2π = ||σA | − |σB || 4πǫ0 1 = (2π) ||σA | − |σB || 4πǫ0 = (2π) × (9 × 109 ) × (2 × 10−9 ) = 36π N C−1 Let E2 be the electric field between the plates A and B. In this region, the electric field EA due to A points in the same direction as the electric field EB due to B. Therefore |E2 | = ||EA | + |EB || 1 ||σA | + |σB || = (2π) 4πǫ0 = (2π) × (9 × 109 ) × ((3 + 1) × 10−9 ) = 72.0π N C−1 Similarly, it can be shown that for E3 |E3 | = |−|EA | + |EB || = |E1 | page 4 of 5 PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS Question 47 An infinite charged sheet has a charge density σ of 0.10 nC mm−2 . How far apart are the equipotential surfaces whose potential differ by 5.00 kV? The field due to an infinite charged sheet is E= σ 2ǫ0 The electric potential is given by the field E and the separation d of the plates: V = Ed For equipotential surfaces whose potential differ by 5.00 kV ∆V = |V2 − V1 | = 5.00 × 103 = d × Therefore the separation d is 2ǫ0 × (5.00 × 103 ) σ 4πǫ0 5.00 × 103 = × 2π 0.1 × 10−9 × 106 = 8.8 × 10−4 m d = page 5 of 5 σ 2ǫ0