Answer to AP Physics C Free Response Practice Question on Electric Field & Potential

In the post dated 2^nd August 2009, the following free-response question for practice was given to you:

A quantity Q of positive charge is placed at the position A (Fig.) on a circular conducting ring of radius R made of thin uniform wire. The system is placed in a region of space where the effect of external charges is negligible. Now, answer the following questions in respect of the above system, assuming the expressions for the electric field and potential due to a point charge:

(a) What is the electric field at the centre of the ring? Justify your answer.

(b):

(i) What is the electric potential at the centre of the ring?

(ii) If the charge placed on the ring is negative, will there be any change in the electric potential at the centre of the ring? Justify your answer.

(c) Derive an expression for the electric potential at a point such as P on the axis of the positively charged ring. (Assume that the axis of the ring is along the x-direction and the centre of the ring is at the origin).

(d):

(i)Using the expression for the electric potential obtained in part (c) above, obtain an expression for the electric field at the point P on the axis of the ring.

(ii) Show that the electric field on the axis is maximum at a distance R/√2 from the centre

(e) Show qualitatively, in a diagram, the nature of variation of the electric field along the axis of the ring, covering both sides of the ring.

As promised, I give below a model answer for the above question:

(a) The electric field at the centre of the ring is zero.

This follows from the definition of the electric field: Electric field at any point is the force per unit positive test charge placed at the point. The test charge placed at the centre of the ring will be repelled equally by the uniformly distributed positive charges on the conducting ring so that the net force on the test charge will be zero.

[Note that the charge Q placed at the position A on the conducting ring will be uniformly distributed immediately throughout the ring].

(b):

(i) The charges on the ring are at the same distance R from the centre of the ring and hence the potential at the centre of the ring is Q/4πε₀R.

(ii) If the charge placed on the ring is negative, the potential will be negative. The potential in this case will be – Q/4πε₀R.

(c) Consider a small quantity (dQ) of charge at a point such as A on the ring. The electric potential dV at P due to this elemental charge is given by

dV = dQ/4πε₀r.

The total potential (V) at P due to all the charges on the ring is given by

V = ∫dV = ∫dQ/4πε₀r = Q/4πε₀r (since all charge elements are at the same distance r from the point P).

(d) The electric field at P due to an element dQ of charge at a point such as A is directed along AP. This field can be resolved into two components: one component along the axis of the ring (axial component) and the other component perpendicular to the axis of the ring (normal component). For every elemental charge at any given point on the ring, there is an equal elemental charge situated diametrically opposite to it. The diametrically opposite element will produce an equal axial component of field in the same direction. But the normal component due to the diametrically opposite elemental charge will be equal in magnitude but opposite in direction. Therefore, all axial components get added where as all normal components get canceled.

The electric field due to the charged conducting ring is therefore directed along the axis of the ring.

The electric field at the point P is given by

E = –grad V.

Since we know that the electric field is along the axis of the ring which is along the x-axis as given in the question, we have

E = –∂V/∂x = –∂/_∂x (Q/4πε₀r) = –∂/_∂x [Q/4πε₀√(R²+x²)]

Or, E = (1/4πε₀) [Qx/(R²+x²)^3/2]

Since the charge on the ring is positive, the field E is along the positive x-direction.

When the electric field is maximum, dE/dx = 0

Therefore, (1/4πε₀)[{(–3/2)(R²+x²)^–5/2 ×2x ×x} + (R²+x²)^–3/2] = 0

Or, 3x² (R²+x²)^–1 = 1

Or, 3x² = R²+x² from which x = R/√2.

(e) The nature of variation of the electric field along the axis of the ring is shown in the adjoining figure in which O represents the centre of the ring. The field has maximum magnitude at points distant R/√2 from the centre and these are located symmetrically on either side of the centre. Positive direction of the field is the positive x-direction even though it is represented by positive y-plot in the figure. The negative direction of the field is the negative x-direction and is represented by negative y-plot in the figure.

You can access all posts related to electrostatics (including multiple choice practice questions) on this site by clicking on the label ‘electrostatics’ below this post.

Electrostatics- AP Physics C Free Response Practice Question on Electric Field & Potential

Essential points to be remembered in respect of electric field and potential were discussed on this blog on 16^th June 2008. You can access that post by clicking here.

All posts related to electrostatics can be accessed by clicking on the label ‘electrostatics’ below this post.

Today I’ll give you a free response practice question involving electric field and potential:

A quantity Q of positive charge is placed at the position A (Fig.) on a circular conducting ring of radius R made of thin uniform wire. The system is placed in a region of space where the effect of external charges is negligible. Now, answer the following questions in respect of the above system, assuming the expressions for the electric field and potential due to a point charge:

(a) What is the electric field at the centre of the ring? Justify your answer.

(b):

(i) What is the electric potential at the centre of the ring?

(ii) If the charge placed on the ring is negative, will there be any change in the electric potential at the centre of the ring? Justify your answer.

(c) Derive an expression for the electric potential at a point such as P on the axis of the positively charged ring. (Assume that the axis of the ring is along the x-direction and the centre of the ring is at the origin).

(d):

(i)Using the expression for the electric potential obtained in part (c) above, obtain an expression for the electric field at the point P on the axis of the ring.

(ii) Show that the electric field on the axis is maximum at a distance R/√2 from the centre

(e) Show qualitatively, in a diagram, the nature of variation of the electric field along the axis of the ring, covering both sides of the ring.

This question carries 15 points and you have 15 minutes for answering it. Try to answer this question. I’ll be back soon with a model answer for your benefit.

AP Physics B & C – Multiple Choice Questions (MCQ) for Practice on Gravitation

Equations to be remembered in respect of the section on gravitation were given in the post dated 9^th May 2008. A few multiple choice practice questions were discussed in the post dated 12^th May 2008 followed by a free response practice question on gravitation in the post dated 15^th May 2008. You can access all posts on gravitation on this site by clicking on the label ‘gravitation’ below this post.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are meant for AP Physics B as well as C:

(1) The speed of an artificial satellite moving in an orbit of radius r around the earth is increased by 41.4%. The satellite will

(a) continue to move in the same orbit

(b) move in an orbit of radius 1.414 r

(c) move in an orbit of radius 0.414 r

(d) move in an orbit of radius 0.586 r

(e) escape into the outer space

The orbital speed v of any satellite is given by

v =√(gr) where g is the acceleration due to gravity at the orbit of radius r.

[You will get this by equating the centripetal force to the gravitational pull:

mv²/r = mg]

The escape velocity (v_escape) of a body at a point distant r from the centre of the earth where the acceleration due to gravity is equal to g is given by

v_escape =√(2gr)

Thus the escape velocity is √2 times the orbital speed.

The speed v of the satellite becomes 1.414 times the initial value on increasing the speed by 41.4%. Now, 1.414 v = √2 v. Therefore the new speed is the escape speed so that the satellite will escape into the outer space.

(2) A launch vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. What is the minimum energy to be spent by the launch vehicle on the satellite so that the satellite will move in a circular orbit of radius 7R? (Gravitational constant = G)

(a) GMm /7R

(b) 6GMm /7R

(c) 6GMm /14R

(d) 13GMm /14R

(e) GMm /14R

The satellite of mass m is initially at rest and therefore its initial energy (E_i) is its gravitational potential energy given by

E_i = – GMm/R

When the satellite is in its orbit of radius 7R, its final energy (E_f) is given by

E_f = – GMm/(2×7R) = – GMm/(14R)

[Note that the total energy of a satellite in an orbit of radius r is – GMm/2r]

The minimum energy required for placing the satellite in its orbit is E_f – E_i given by

E_f – E_i = – [GMm/(14R)] – [– GMm/R]

= (GMm/R) – [GMm/(14R)]

= 13GMm /14R

(3) A small body is projected vertically up with a speed equal to half the escape speed from the earth’s surface. If the radius of the earth is R, what is the maximum height (h) reached by the body? (Neglect air resistance).

(a) R/3

(b) R/2

(c) R

(d) 2R

(e) 5R

When the body projected vertically reaches the maximum height, its speed will be zero since its initial kinetic energy us used up in increasing its gravitational potential energy. Therefore we have

½ m (v_e/2)² = [–GMm /(R+h)] – [–GMm /R] where m is the mass of the body, M is the mass of the earth, v_e is the escape speed from the earth’s surface and G is the gravitational constant.

[The first term on the right hand side of the above equation is the gravitational potential energy of the body at height h. The second term is the gravitational potential energy of the body on the surface of the earth].

Since v_e = √(2GM/R), the above equation becomes

GM /4R = (GM /R) –[GM /(R+h)]

Or, GM /4R = GMh /[R(R+h)

Therefore, ¼ = h/(R+h) from which h = R/3

(4) The orbit of a planet moving around the sun is elliptical with the sun S at one focus of the ellipse. If the orbital speeds of the planet while at points A and B (Fig.) are V₁ and V₂ and the distances SA and SB are d₁ and d₂ respectively, V₁/ V₂ is equal to

(a) √(d₁ /d₂)

(b) √(d₂ /d₁)

(c) d₁ /d₂

(d) d₂ /d₁

(e) (d₂ + d₁) /(d₂ – d₁)

To obtain the answer you may apply Kepler’s law of areas (in respect of plnetary motion), which says that the straight line joining the sun to the planet sweeps equal areas in equal intervals of time. The adjoining figure shows path lengths CD and EF traced by the planet in equal intervals (say, ∆t) and hence the areas CADS and EBFS must be equal. If ∆t is sufficiently small, CD and EF will become straight line segments and the areas will become triangular. Equating the areas of these triangles, we have

½ ×CD×AS = ½ ×EF×BS

Or, ½ × V₁ ∆t × d₁ = ½ × V₂ ∆t × d₂

Therefore, V₁/ V₂ = d₂ /d₁

[You can easily get the answer from angular momentum conservation:

m V₁ d₁ = m V₂ d₂

where m is the mass of the planet. This gives V₁/ V₂ = d₂ /d₁]

(5) We know that the gravitational force between two point masses is inversely proportional to the square of the distance between them. If the gravitational force between two point masses were (let us imagine) inversely proportional to the n^th power of the distance between them, what would be the relation between the orbital period (T) of a planet around the sun and the mean distance (r) of the planet from the sun?

(a) T² α r²ⁿ

(b) T² α rⁿ

(c) T² α r²ⁿ⁺¹

(d) T² α r⁽ⁿ⁺¹⁾

(e) T² α r^{(n – 1)}

For simplicity, let us assume that the orbit is circular. (Circular orbit is only a special case of the general elliptical orbit and the relation between T and r will be unchanged). The centripetal force required for the circular motion is supplied by the gravitational pull so that we have

mrω² = GMm/rⁿ where m is the mass of the planet, ω is its orbital angular velocity, M is the mass of the sun and G is the gravitational constant.

The above equation gives ω² = GM/r⁽ⁿ⁺¹⁾

Since ω = T/2π, we have

(T/2π)² = GM/r⁽ⁿ⁺¹⁾

Therefore, T² α r⁽ⁿ⁺¹⁾

You will find some useful multiple choice questions on gravitation (with solution) at physicsplus.

AP Physics B - Multiple Choice Questions on Fluid Mechanics

Essential things you need to remember in fluid mechanics were discussed in the post dated 3^rd December 2007. A few multiple choice questions (with solution) in this section were discussed in the posts dated 4^th and 5^th December 2007. You can access all posts related to fluid mechanics on this site by clicking on the label ‘fluid mechanics’ below this post.

Today we will discuss a few more multiple choice questions on fluid mechanics:

(1) A water storage tank has a square hole of side 2 cm at its bottom. A plumber, unaware of the hole, admits water into the tank at a constant rate of √2 litre per second. Up to what height can water remain in the tank? (g = 10 ms^–2)

(a) 12.25 m

(b) 8.25 m

(c) 2.45 m

(d) 1.25 m

(e) 0.625 m

If h is the height up to which water can remain in the tank, the velocity v of water flowing out through the hole in the steady state is given by

v = √(2gh)

The volume of water flowing out in the steady state is av where a is the area of the hole.

This must be equal to the volume of incoming water which is given as √2 litre = √2 ×10^–3

m³. Thus we have

a√(2gh) = √2 ×10^–3

Here a = 4 cm² = 4×10^–4 m².

Substituting, 4×10^–4×√(2×10×h) = √2 ×10^–3

Or, √(10h) = 2.5 from which h = 6.25/10 = 0.625 m.

(2) A piece of granite floats at the interface of mercury and water contained in a beaker (Fig.). If the densities of granite, water and mercury are ρ, ρ₁ and ρ₂ respectively, the ratio of the volume of granite in water to the volume in mercury is

(a) (ρ₂ – ρ) /(ρ – ρ₁)

(b) (ρ₂ + ρ) /(ρ+ ρ₁)

(c) ρ₁ ρ₂ /ρ

(d) ρ₁ /ρ₂

(e) ρ₂ /ρ₁

The weight of a floating body is equal to the weight of the displaced fluid. If V and v represent the total volume of the piece of granite and volume of granite in water respectively, we have

V ρg = v ρ₁g + (V – v) ρ₂g

Or, v(ρ₁ – ρ₂) = V(ρ – ρ₂)

Therefore, v/V = (ρ – ρ₂) /(ρ₁ – ρ₂)

The ratio required in the question is v/ (V–v) and is given by

v/ (V–v) = (ρ – ρ₂) /[(ρ₁ – ρ₂) –(ρ – ρ₂)]

Or, v/ (V–v) = (ρ – ρ₂) /(ρ₁ – ρ) = (ρ₂ – ρ) /(ρ – ρ₁)

(3) The pressure of water at the bottom of a water tank exceeds the atmospheric pressure by 10⁴ pascal. The velocity of efflux of water through an orifice at the bottom of the water tank will be (g = 10 ms^–2)

(a) √5 ms^–1

(b) √10 ms^–1

(c) √15 ms^–1

(d) √20 ms^–1

(e) √30 ms^–1

If h is the height of water column in the tank, we have

hρg = 10⁴

Since the density (ρ) of water is 1000 kg m^–3,

h×1000×10 = 10⁴ from which h = 1m.

The velocity of efflux (Torricelli’s theorem) is √(2gh) = √(2×10×1) = √20 ms^–1

(4) In a wind tunnel the flow speeds (of air) on the upper and lower surfaces of the wing of a model airplane are v₁ and v₂ respectively (v₁> v₂). If the wing area is A and the density of air is ρ, the lift on the wing is

(a) ρ (v₁² – v₂²)A

(b) ½ ρ (v₁ – v₂)A

(c) ½ ρ (v₁² – v₂²)A

(d) ½ (v₁² – v₂²)A /ρ

(e) ρ (v₁ – v₂)A

If P₁ and P₂ are the pressures on the upper and lower surfaces of the wing we have, by Bernoulli’s theorem,

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

[We have ignored the small height difference between the top and bottom of the wing so that the gravitational potential energy is treated as constant].

The pressure at the top is less than the pressure at the bottom since v₁> v₂.

The lift on the wing is (P₂ – P₁)A = ½ ρ (v₁² – v₂²)A

You will find similar useful multiple choice questions with solution here as well as here at Physicsplus.