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Potential Difference - Physics with Calculus - Solved Exam, Exams of Physics

Acharya Nagarjuna UniversityPhysics

This is the Solved Exam of Physics with Calculus which includes Voltage Across Thicker Section, Downward Electric Field, Unpolarized Light, Intensity of Light, Gaussian Surface, Shortened Heating Element etc. Key important points are: Potential Difference, Kinetic Energy, Dipole with Dipole Moment, Constant Electric Field, Equivalent Capacitance of Arrangement, Gaussian Surface, Distribution of Charge, Surface Charge Density

Typology: Exams

2012/2013

Uploaded on 02/23/2013

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Solution to Exam 1
Paul Avery, Arthur Hebard
PHY2049, Fall 2003
Sep. 24, 2003
1. Two parallel conducting plates are brought to a potential diļ¬€erence of 4000 V and a
small object of mass 4 mg and charge 0.2ĀµC accelerates from rest at one plate. With
what speed in m/s will it reach the other plate?
Solution: The work done by the electric ļ¬eld is qV ,whereqis the charge and Vis
the potential diļ¬€erence. Thus the kinetic energy is K=qV . Solving 1
2mv2=qV for
the velocity vgives v=ī˜2qV/m, yielding v= 20m/s.
2. Consider a dipole with dipole moment poriented at an angle Īøwith respect to a
constant electric ļ¬eld E.AtwhatangleĪøin degrees is the torque Ļ„a maximum, and
what is the magnitude of Ļ„for this orientation?
Solution: The torque is maximum when the dipole is oriented perpendicular to the
electric ļ¬eld. Thus Īø=90
ā—¦and Ļ„=pE.
3. Five capacitors, each of capacitance C, are connected as shown in the ļ¬gure. What is
the equivalent capacitance of the arrangement?
Solution: Itā€™s best to do this in stages. Capacitors add in parallel, so the arrangement
can be immediately simpliļ¬ed to 2C,Cand 2Cin series. The equivalent capacitance
can be found from Ceq =1/2C+1/C +1/2C, giving Ceq =C/2.
4. A Gaussian surface, in the shape of a rectangular box, can be used to ļ¬nd the electric
ļ¬eld from a distribution of charge. What symmetry must be exhibited by the charge
distribution for this to work?
Solution: The charge must exhibit planar symmetry so that the electric ļ¬eld lines are
perpendicular to at least one side of a rectangular box.
5. An electron is located a distance x0to the right of the midpoint of a vertically oriented
electric dipole as shown in the diagram. The direction of the force on the electron is
1
pf3

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Solution to Exam 1

Paul Avery, Arthur Hebard PHY2049, Fall 2003 Sep. 24, 2003

  1. Two parallel conducting plates are brought to a potential difference of 4000 V and a small object of mass 4 mg and charge 0.2Ī¼C accelerates from rest at one plate. With what speed in m/s will it reach the other plate? Solution: The work done by the electric field is qV , where q is the charge and V is the potential difference. Thus the kinetic energy is K = qV. Solving 12 mv^2 = qV for the velocity v gives v =

āˆš 2 qV /m, yielding v = 20m/s.

  1. Consider a dipole with dipole moment p oriented at an angle Īø with respect to a constant electric field E. At what angle Īø in degrees is the torque Ļ„ a maximum, and what is the magnitude of Ļ„ for this orientation? Solution: The torque is maximum when the dipole is oriented perpendicular to the electric field. Thus Īø = 90ā—¦^ and Ļ„ = pE.
  2. Five capacitors, each of capacitance C , are connected as shown in the figure. What is the equivalent capacitance of the arrangement?

Solution: Itā€™s best to do this in stages. Capacitors add in parallel, so the arrangement can be immediately simplified to 2C, C and 2C in series. The equivalent capacitance can be found from Ceq = 1/ 2 C + 1/C + 1/ 2 C, giving Ceq = C/2.

  1. A Gaussian surface, in the shape of a rectangular box, can be used to find the electric field from a distribution of charge. What symmetry must be exhibited by the charge distribution for this to work? Solution: The charge must exhibit planar symmetry so that the electric field lines are perpendicular to at least one side of a rectangular box.
  2. An electron is located a distance x 0 to the right of the midpoint of a vertically oriented electric dipole as shown in the diagram. The direction of the force on the electron is

x 0

e āˆ’

Solution: The dipole is oriented so that the negative charge is at the base of the arrow and the positive charge is at the tip. The net horizontal is easily seen to cancel and each charge gives a positive upward component to the negatively charged electron. Thus the total force on the electron is upward.

  1. A conducting spherical shell has inner radius r and outer radius R, as shown in the diagram. A point charge of āˆ’q is located at the center of the sphere and a charge of +Q is placed on the conducting shell. What is the surface charge density at the outer surface of the conducting shell?

R

  • Q

āˆ’ q

r

Solution: The charge in the shell will distribute itself on the inner and outer surfaces in such a way that the field inside the conductor is zero. Using Gaussā€™ law, we draw an imaginary surface inside the shell (radius between r and R). Since the total flux through the surface is zero, the total enclosed charge is zero and therefore the charge on the inner surface must be q (to cancel the charge āˆ’q in the center). Thus the total charge on the outer surface is Qāˆ’q. The outer charge density is therefore (Qāˆ’q)/ 4 Ļ€R^2.

  1. Capacitor C 1 is charged by the voltage V when the switch is in position 1 as shown in the figure. The switch is then moved to position 2 and the charge redistributes between the two capacitors. After this redistribution, what is the charge on C 1?

V

C 1 C 2