AP Physics B and C- Electrostatics: Additional Multiple Choice Practice Questions on Field and Potential

As promised in the post dated 20^th June 2008, we will discuss a few more multiple choice questions on electric field and potential.

The following questions are meant for AP Physics B as well as AP Physics C:

(1) Three positive charges each of magnitude q are arranged symmetrically along the circumference of a circle of radius r. The electric field at the centre of the circle has magnitude

(a) 3q /4πε₀r²

(b) √3q /4πε₀r²

(c) q/4πε₀r²

(d) (1+√2)q/4πε₀r²

(e) zero

The electric field at any point is the force acting on unit positive charge. The three charges arranged symmetrically along the circle will exert (on unit positive charge placed at the centre) forces of the same magnitude with their lines of action inclined at 120º. The resultant force is therefore zero. Hence the field at the centre of the circle is zero.

(2) A thin ring of radius R is uniformly charged with a total amount of positive charge Q. The electric potential at a point P on the axis of the ring at distance x from the centre of the ring is

(a) Q /4πε₀√( r²+x²)

(b) Qr /2ε₀√( r²+x²)

(c) Q /4πε₀ ( r²+x²)

(d) Qr /4πε₀√( r²+x²)

(e) Qx/4πε₀√( r²+x²)

The distance of point P from the charges on the ring is √( r²+x²) so that the electric potential at P is Q /4πε₀√( r²+x²) given in option (a).

(3) Two identical conducting spheres, each of radius R, carry charges +Q and –Q/2 respectively. When the distance between their centres is d (d>2R), the attractive force between them is F. The spheres are now brought into contact and then separated to the initial distance d. The force between the spheres is now

(a) attractive and of magnitude F/4

(b) repulsive and of magnitude F/4

(c) attractive and of magnitude F/8

(d) repulsive and of magnitude F/8

(e) zero

The initial attractive force between the charged spheres is F = (1/4πε₀) (Q²/2d²)

When the spheres are brought into contact they share the charges. The total charge on the system of the two spheres is the sum of the initial charges and is equal to +Q/2. Since the spheres are identical, they share the charges equally so that the charge on each sphere is +Q/4.

The final force between the spheres is therefore repulsive and of magnitude (1/4πε₀) (Q²/16d²) = F/8.

The following practice questions are for AP Physics C aspirants only:

(1) Two identical thin non-conducting spherical shells of radius R carry equal negative charges of magnitude Q sprayed uniformly on their surfaces. The distance AB between their centres is 4R. The electric field at a point P (Fig.) at distance R/2 from the centre A of the first sphere is

(a) Q/64πε₀R² directed towards right

(b) Q/64πε₀R² directed towards left

(c) Q/49πε₀R² directed towards left

(d) Q/49πε₀R² directed towards right

(e) zero

The electric field at P is the resultant of the fields due to the two shells. But since the point P is inside the first shell (with centre at A), the field due to the first shell is zero. Therefore the field at P is solely due to the second shell. The distance of the centre of the second shell from P is 4R – (R/2) = 7R/2. Therefore the magnitude of the field at P is given by

E = (1/4πε₀)[Q/(7R/2)²] = Q/49πε₀R²

Since the charge on the sphere is negative, the field at P is directed towards right [Option (d)].

(2) If the spherical shells in the above question are made of a conductor, the electric field at P will be

(a) Q/64πε₀R² directed towards right

(b) Q/64πε₀R² directed towards left

(c) Q/49πε₀R² directed towards left

(d) Q/49πε₀R² directed towards right

(e) zero

Since the point P is inside the first shell, the field at P due to the charges on the first shell is zero as in the above question. The field at P due to the charges on the second shell also is zero since the conducting shell provides electrostatic screening of the electric field produced by the second shell. Therefore the correct option is (e).

(3) A thin walled cubical box of side 10 cm has a constant potential of 4 V on its surface. There are no charges inside the box. The potential at the centre of the box is

(a) 2V

(b) 4V

(c) zero

(d) 8V

(e) 2V/3

Since there are no charges inside the box, there cannot be any potential gradient on that account. Since the potential everywhere on the surface of the box is 4 V, there is no potential gradient inside the box on this account also. To put this in a different manner, there is no electric field inside the box and hence the potential at the centre is 4 V itself.

(4) A charge Q is divided into two parts as q and (Q – q) so that for any given separation between q and (Q – q), the electrostatic interaction between q and (Q – q) is maximum. The ratio Q/q is

(a) 8

(b) 6

(c) 4

(d) 3

(e) 2

The electrostatic force between the two parts is repulsive and is given by

F = q(Q – q) /4πε₀r²

This will be maximum when dF/dq = 0

Therefore, –2q + Q = 0 from which q = Q/2.

The ratio Q/q = 2

** ** ** ** ** ** ** ** ** ** ** **

You know that a soap bubble has an excess of pressure inside because of surface tension. What will happen to the excess of pressure on charging the bubble? Because of the repulsion between the charges, the bubble will expand, thereby reducing the excess of pressure.

AP Physics B and C- Electrostatics: Multiple Choice Questions on Field and Potential

In the post dated 16^th June 2008, the essential points you have to note under electric potential and electric field in Electrostatics were discussed. As promised, we will discuss some multiple choice questions in this section.

The following questions are meant for AP Physics B as well as AP Physics C aspirants:

(1) For transferring a charge +Q coulomb from point A to point B separated by a distance d, the work required to be done by an external agency is –W joule. If the potential of A is V volt, the potential of B will be (in volt)

(a) V + (W/Q)

(b) V + (Q/Wd)

(c) V + (W/Qd)

(d) V – (W/Qd)

(e) V – (W/Q)

The potential difference between two points is the work done in moving unit charge (coulomb) from one point to the other. So the potential difference between A and B is W/Q volt. The work done is negative here because the point B is at a lower potential so that the external agency need not do any positive work. The charge will move to B by itself under the attractive force towards B (and hence the work done by the external agency is negative).

The potential of B is therefore equal to V – (W/Q) volt.

The separation d between the points serves only as a distraction in this question.

(2) A point charge q is placed at the centre of a skeleton cube made of thin non-conducting wire. The electric flux passing through one face of the cube is

(a) q/6ε₀

(b) q/24ε₀

(c) q/8ε₀

(d) q/8πε₀

(e) q/6πε₀

Total electric flux from a charge q is q/ε₀. Since the charge q is situated at the centre of the cube (fig.), all the six faces of the cube have equal share of the flux so that the flux through one face is q/6ε₀.

(3) A particle of mass m and charge q is projected with speed v towards a stationary particle of mass M and charge Q. The projected particle reaches a distance of closest approach d and retraces its path. If the same particle is projected with speed v/2, the distance of closest approach will be

(a) d

(b) d/2

(c) 2d

(d) d/4

(e) 4d

The distance of closest approach d is obtained by equating the kinetic energy of the projected particle to its electrostatic potential energy (since the electrostatic potential energy is increased at the cost of kinetic energy). Therefore we have

½ mv² = (1/4πε₀)Qq/d

Evidently, if the velocity v of the projectile is halved, the distance of closest approach will be quadrupled. The correct option is 4d.

The following questions are meant for AP Physics C only:

(1) A charge +4q is located on the X-axis of a Cartesian coordinate system at x = – 2a as shown. Another charge +(q/4) is located on the X-axis at x = a. Where on the Y-axis should a third charge –3q be placed so that the resultant electric field at the origin O has equal positive X and Y components?

(a) y = a

(b) y = 2a

(c) y = –3a

(d) y = –2a

(e) y = (¾)a

The resultant field (E) at the origin due to the charges +4q and +(q/4) is given by

(1/4πε₀)[4q/(2a)² – (q/4)/a²] = (1/4πε₀) (3q/4a²)

This field is directed along the positive X-direction.

The field produced by the third charge –(3q) should have the same magnitude as that of the above field and should be directed along the positive Y-direction so as to ensure equal positive X and Y components for the resultant field (due to all the three charges) at the origin.

Therefore, the charge –3q has to be placed on the positive Y-axis at distance y given by

(1/4πε₀) ×3q/y² = (1/4πε₀) (3q/4a²)

This gives y = 2a.

(2) A thin non conducting spherical shell of radius R has total charge Q spread uniformly over its surface. Which of the following graphs represents the variation of electric field E(r) (due to the shell) with the distance r from the centre of the shell?

The correct option is (c) since the field inside a uniformly charged thin spherical shell is zero and the fields on the surface of the shell (r = R) and for points outside (r >R) are respectively (1/4πε₀)Q/R ² and (1/4πε₀)Q/r ².

(3) Two concentric thin spherical shells of radii R and r (R > r) share a charge Q so that their surface charge densities are the same. The electric potential at the common centre of the shells is (1/4πε₀ = k)

(a) kQ/(R – r)

(b) kQ/(R + r)

(c) kQ(R – r)/(R² + r²)

(d) kQ(R + r)/(R² + r²)

(e) kQ/(R

The potential at the centre due to the larger and smaller shells are respectively (1/4πε₀) q₁/R and (1/4πε₀) q₂/r where q₁ = 4πR²σ and q₂ = 4πr²σ. Here σ is the surface charge density.

Since the total charge in the system is Q, we have

Q = q₁ + q₂ = 4πσ(R² + r²) from which σ = Q/[4π(R² + r²)]

The total potential at the centre is the sum of the above potentials and is given by

V = (1/4πε₀)[(4πR²σ/R) + (4πr²σ/r)]

= (1/4πε₀) × 4πσ (R + r) = σ (R + r)/ε₀

Substituting for σ, V = (1/4πε₀) ×Q (R + r)/(R² + r²) = kQ(R + r)/(R² + r²)

In the next post we will discuss some more multiple choice questions. Free response questions will be discussed after that.

It is high time that the ideal of success should be replaced by the ideal of service.

–Albert Einstein 

AP Physics B and C – Electrostatics- Field & Potential- Equations to be remembered

We will discuss the essential points to be remembered in electrostatics under electric field and potential. Gauss’s law as well as field and potential due to charge configurations other than those involving point charges are meant for AP Physics C only.

(1) Coulomb’s Law: The electrostatic force between two point charges q₁ and q₂ is proportional to the product q₁q₂ and inversely proportional to the square of the distance r between them.

This is expressed mathematically as,

F = k (q₁q₂) /r²

where the constant k = 1/4πε₀ which is very nearly equal to 9×10⁹ Nm²C^–2. The constant ε₀ is the permittivity of free space. The equation for electrostatic force when the charges are in free space is therefore written as

F = (1/4πε₀) (q₁q₂) /r²

[If the charges are placed in a medium of relative permittivity ε_r, the above equation for electrostatic force will become F = (1/4πε₀ε_r) (q₁q₂) /r²].

In vector form, the equation for electrostatic force when the charges are in free space is

F₂₁ = k (q₁q₂) ř₂₁/ r₂₁²

In this form F₂₁ is the force (vector) on q₂ due to q₁, ř₂₁ is a unit vector in the direction from q₁ to q₂ and r₂₁ is the distance between q₁ and q₂.

(2) The electric field E at any point is the force on a small positive test charge q placed at the point divided by the magnitude of the test charge:

E = F/q

(3) Electric field (magnitude E) due to a point charge Q at distance r is given by

E = (1/4πε₀)Q/r². It is radially outwards from q, if q is positive and radially inwards if q is negative.

(4) Electric field due to an electric dipole at a point on its axis distant r from the centre of the dipole s given by

E = (1/4πε₀) 2pr/(r² – a²)² where p is the dipole moment vector of magnitude 2qa directed from –q to +q.

[You should remember that 2a is the separation between the charges –q and +q constituting the dipole].

The magnitude of the electric field due to an electric dipole at a point on its axis distant r from the centre of the dipole s given by

E = (1/4πε₀) 2pr/(r² – a²)² where p = 2qa.

At a distance r large compared to the dipole length 2a, the field is

E = (1/4πε₀) 2p/r³

(5) Electric field due to an electric dipole at a point in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) distant r from the centre of the dipole s given by

E = (1/4πε₀) (–p)/(r² +a²)^3/2

The field in the equatorial plane is antiparallel to the dipole moment vector as indicated by –p in the above expression.

The magnitude of the electric field due to an electric dipole at a point in its equatorial plane distant r from the centre of the dipole is given by

E = (1/4πε₀) p/r³

[Note that the expression for the electric field due to an electric dipole is similar to the expression for the magnetic field due to a magnetic dipole, which is obtained by replacing (1/4πε₀) with (μ₀/4π) and the electric dipole moment p with the magnetic dipole moment m].

(6) Torque (τ) on an electric dipole of moment p placed in a uniform electric field E is given by

τ = p×E

Therefore, τ = pE sinθ where θ is the angle between p and E

[Remember that the dipole moment vector has magnitude 2qa and its direction is from –q to +q].

(7) Gauss’s Law (Gauss Theorem): The flux (Ф) of electric field through any closed surface S is 1/ε₀ times the total charge enclosed by the surface S.

Ф = ∫E.dS = Q/ε₀ where the integration is over the closed surface S which encloses a total charge Q.

(8) Electric fields due to some symmetric charge configurations:

(i) Electric field (magnitude E) due to a spherical conductor carrying charge Q at a point outside the sphere at distance r from the centre of the sphere is given by

E = (1/4πε₀)Q/r ².

The field on the surface of the spherical conductor is

E = (1/4πε₀)Q/R ² where R is the radius of the sphere.

As far as points outside the sphere and on the sphere are concerned, the electric field due to a charged spherical conductor has values as though the entire charge is concentrated at the centre.

(ii) Electric field at any point inside a conductor carrying charge Q is zero since the charge resides only on the surface of the conductor.

(iii) Electric field (magnitude E) due to a uniformly charged thin spherical shell carrying charge Q is given by

E = (1/4πε₀)Q/r ² at a point distant r from the centre, outside the shell.

On the surface of the shell, E = (1/4πε₀)Q/R ² where R is the radius of the shell.

Obviously, at any point inside a uniformly charged thin spherical shell carrying charge Q, the field is zero.

(iv) Electric field due to a spherical distribution of charges with a uniform volume charge density ρ:

At a point outside the sphere at distance r from the centre of the sphere the field is given by

E = ρR³/3ε₀r²), which follows by substituting Q = (4/3) πR³ρ in the expression E = (1/4πε₀)Q/r ².

At a point on the surface of the sphere the field is given by

E = ρR/3ε₀, which follows by substituting r = R in the above expression.

At a point inside the sphere at distance r from the centre of the sphere the field is given by

E = ρr/3ε₀), which follows by substituting Q = (4/3) πr³ρ in the expression E = (1/4πε₀)Q/r ².

(v) Electric field at distance r from an infinitely long straight uniformly charged wire with linear charge density λ is

E = λ/2πε₀r

The field is perpendicular to the wire.

(vi) Electric field due to a uniformly charged infinite plane sheet with uniform surface charge density σ:

E = σ/2ε₀, which is independent of the distance of the point from the surface.

The field is directed normal to the surface.

(9) Electrostatic potential (V) at a point is the work done (by an external agency) in bringing unit positive charge from infinity to that point.

By convention, the potential at infinity is taken as zero.

(i) Electrostatic potential due to a point charge Q at distance r is given by

V = (1/4πε₀)Q/r

(ii) Potential at a point P due to a system of charges q₁, q₂, q₃,….etc. at distances r₁, r₂, r₃,….etc. is given by

V = (1/4πε₀)[(q₁/r₁) + (q₂/r₂) + (q₃/r₃) +…etc.]

(iii) Electrostatic potential due to an electric dipole at distance r (which is very large compared to the dipole length 2a) from the centre of the dipole is given by

V = p cosθ /4πε₀r²

where p is the magnitude of the dipole moment and θ is the angle between the dipole moment vector and the line joining the dipole to the point P where the potential is measured.

[Note that potential is a scalar quantity. In terms of the dipole moment vector p and the position vector r of the point P, taking the centre of the dipole as the origin, the potential of the dipole can be written as V = = p.r/4πε₀r³. Or, V = = p.ř/4πε₀r² where ř is a unit vector along the position vector r].

(iv) Electric potential due to a spherical conductor of radius R carrying charge Q:

At a point outside the sphere at distance r from the centre of the sphere, the potential is given by

V = (1/4πε₀)Q/r

At a point on the surface of the sphere the potential is given by

V = (1/4πε₀)Q/R

At all points inside the sphere, the potential is the same as the value at the surface given by

V = (1/4πε₀)Q/R

(v) Electric potential at any point due to a uniformly charged thin spherical shell carrying charge Q is the same as for a charged spherical conductor (given above).

(10) Electrostatic potential energy of a charge q at a point where the electric potential is V is Vq.

(11) Electrostatic potential energy (U) of a system of two point charges q₁ and q₂ separated by a distance r is given by

U = (1/4πε₀)q₁q₂ /r

Note that U is the work done (by external agency) in reducing the separation between the charges from infinity to r. Its value will be positive or negative, depending on whether the product q₁q₂ is positive or negative.

If there are more than two charges, there will be as many terms in the expression for electrostatic potential energy as there are independent pairs of charges. Thus, if there are three point charges q₁, q₂ and q₃ there will be three terms:

U = (1/4πε₀)[(q₁q₂ /r₁₂) + (q₁q₃ /r₁₃) + (q₂q₃ /r₂₃)] where r₁₂ is the distance between charges q₁ and q₂, r₁₃ is the distance between charges q₁ and q₃ and r₂₃ is the distance between charges q₂ and q₂₃.

If there are four point charges, there will be six terms in U.

We will discuss questions in this section in the next post.

Meanwhile, find some useful multiple choice questions from electrostatics here as well as here