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