ch24 - CHAPTER 24 Electric Potential 1* A uniform electric...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 24 Electric Potential 1* · A uniform electric field of 2 kN/C is in the x direction. A positive point charge Q = 3 μ C is released from rest at the origin. ( a ) What is the potential difference V (4 m) – V (0)? ( b ) What is the change in the potential energy of the charge from x = 0 to x = 4 m? ( c ) What is the kinetic energy of the charge when it is at x = 4 m? ( d ) Find the potential V ( x ) if V ( x ) is chosen to be ( d ) zero at x = 0, ( e ) 4 kV at x = 0, and ( f ) zero at x = 1 m. ( a ) Use Equ. 24-2 b ; ? V = – E ?x ( b ) ? U = q ? V ( c ) Use energy conservation ( d ), ( e ), ( f ) Use Equ. 24-2 b ? V = –8 kV ? U = –24 mJ K = 24 mJ ( d ) V ( x ) = –(2 kV/m) x ; ( e ) V ( x ) = 4 kV – (2 kV/m) x ; ( f ) V ( x ) = 2 kV –(2 kV/m) x 2 · An infinite plane of surface charge density s = +2.5 μ C/m 2 is in the yz plane. ( a ) What is the magnitude of the electric field in newtons per coulomb? In volts per meter? What is the direction of E for positive values of x ? ( b ) What is the potential difference V b V a when point b is at x = 20 cm and point a is at x = 50 cm? ( c ) How much work is required by an outside agent to move a test charge q 0 = +1.5 nC from point a to point b ? ( a ) E x = s /2 e 0 ( b ) ? V = – E x ? x ( c ) W = ? U = q ? V E x = (2.5 × 10 –6 /2 × 8.85 × 10 –12 ) N/C = 141 kN/C V b V a = (–141 × –0.3) kV = 42.4 kV W = (42.4 × 10 3 × 1.5 × 10 –9 ) J = 63.6 μ J 3 · Two large parallel conducting plates separated by 10 cm carry equal and opposite surface charge densities such that the electric field between them is uniform. The difference in potential between the plates is 500 V. An electron is released from rest at the negative plate. ( a ) What is the magnitude of the electric field between the plates? Is the positive or negative plate at the higher potential? ( b ) Find the work done by the electric field on the electron as the electron moves from the negative plate to the positive plate. Express your answer in both electron volts and joules. ( c ) What is the change in potential energy of the electron when it moves from the negative plate to the positive plate? What is its kinetic energy when it reaches the positive plate? ( a ) E is uniform; E = ? V /? x (b) W = q ? V (c) ? U = – W ; K = W E = 500/0.1 V/m = 5 kV/m; high V at the positive plate W = 500 × 1.6 × 10 –19 J = 8 × 10 –17 J = 500 eV ? U = –500 eV; K = 500 eV
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 24 Electric Potential 4 · Explain the distinction between electric potential and electrostatic potential energy. The difference is analogous to that between gravitational potential and gravitational potential energy. Electrostatic potential refers to the difference in potential energy per unit charge between the point of interest and an arbitrary reference point; electrostatic potential energy is the potential energy of a charge with respect to an arbitrary zero. 5* ·
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 20

ch24 - CHAPTER 24 Electric Potential 1* A uniform electric...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online