AP Physics B & C - Electrostatics - Multiple Choice Practice Questions

“Iron rusts from disuse, stagnant water loses its purity and in cold weather becomes frozen; so does inaction sap the vigors of the mind.”

– Leonardo da Vinci

 

Questions on electrostatics were discussed on many occasions on this site. You can access them by clicking on the label ‘electrostatics’ below this post or by trying a search for ‘electrostatics’ using the search box provided on this page. Today we shall discuss a few more questions in this section.

(1) Right angled triangle ABC is located in a uniform electric field E. Sides AC and BC have lengths 0.5 m and 0.3 m respectively and BC is at right angles to the electric field lines (Fig.). If the electric potential difference between A and C is 80 V, what is the magnitude of the electric field E?

(a) 80 V/m

(b) 160 V/m

(c) 200 V/m

(d) 100 V/m

(e) 16 V/m

The electric potential at the point B is the same as that at the point C (since the straight line BC is at rjght angles to the direction of the uniform electric field E). Therefore the potential difference between points A and B is 80 V. The electric field E is directed along AB and hence the magnitude of E is (80 V)/(0.4 m) = 200 V/m.

[AB = √(AC² – BC²) = √(0.5² – 0.3²)  = 0.4 m]

(2) In the above question, what is the component of electric field along the direction AC?

(a) 80 V/m

(b) 160 V/m

(c) 200 V/m

(d) 100 V/m

(e) 40 V/m

The potential difference between the points A and C is 80 V and the distance between these points is 0.5 m. Therefore, the component of electric field along the direction AC is (80 V)/(0.5 m) = 160 V/m.

(3) A cube of side a has a charge Q at its centre (Fig.). What is the electric flux through one face of the cube?

(a) Q/ε₀a

(b) Qa/ε₀

(c) Q/6ε₀

(d) Q/8ε₀

(e) Qε₀/6

The electric flux over a closed surface, according to Gauss theorem, is Q/ε₀ where Q is the net charge enclosed by the surface and ε₀ is the permittivity of free space.

[You can understand the above even without knowing Gauss law. You know (from inverse square law) that the electric field at distance r from a point charge Q is Q/4πε₀r². The electric field at any point is the electric flux through unit area held with the plane of the area perpendicular to the electric field lines. Imagine a spherical surface of radius r such that the charge Q is at the centre. Since the area of the spherical surface is 4πr² the electric field at any point on the surface must be Ф/4πr² where Ф is the total electric flux produced by the charge Q. Therefore we have

            Q/4πε₀r² = Ф/4πr²

This gives Ф = Q/ε₀ as stated in Gauss law.]

Since the charge Q is at the centre of the cube, all the six faces of the cube receive equal electric flux so that the flux through one face of the cube is Ф/6 = Q/6ε₀

(4) The effective capacitance between terminals A and B in the network shown in the adjoining figure is

(a) 16 μF

(b) 8 μF

(c) 6 μF

(d) 16/3 μF

(e) 8/3 μF

Since the capacitors C₁, C₂, C₃ and C₄ are of values satisfying the condition C₁/C₂ = C₃/C₄, the junction of C₁ and C₂ is at the same potential as the junction of C₃ and C₄ (as in the case of a balanced Wheatstone brige). Therefore the capacitors C₅ and C₆ connected across the diagonal of the bridge have no effect and can be ignored.

The network therefore simplifies to the series combination of C₁ and C₂ connected in parallel with the series combination of C₃ and C₄.

Series combined value of C₁ and C₂ = (3×6)/(3+6) = 2 μF

Series combined value of C₃ and C₄ = (1×2)/(1+2) = 2/3 μF

Therefore the effective capacitance between terminals A and B = 2 μF + (2/3) μF = 8/3 μF

The capacitors C₁ and C₄ in the above question are short circuited and the circuit then gets modified as shown in the figure. What is the effective capacitance between the terminals A and B in this situation?   

(a) 11 μF

(b) 9 μF

(c) 8 μF

(d) 7.5 μF

(e) 6.5 μF

Have a careful look at the circuit. You will find that one plate of C₂, C₃ and C₅ is connected to terminal A. The other plate of C₂ as well as C₃ is connected to terminal B while that of C₅ is connected 6through C₆ to terminal B. The series combination of C₅ and C₆ gives an effective capacitance of 1μF. Thus we have three capacitances 1μF, 1μF and 6μF connected in parallel across the terminals A and B.

Therefore the effective capacitance between terminals A and B in this case is 1μF + 1μF + 6μF = 8μF.