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POTENTIAL POINT IN APPLIED PHYSICS, Slides of Physics

National University of Science and Technology (NUST)Physics

The potential computations and electric field of charged spherical and cylindrical conductors. It explains the equations for potential and electric field, as well as the concept of equipotential surfaces and the gradient of potential. The document also provides an example calculation for the maximum potential a terminal of an electrical device can bear.

Typology: Slides

2021/2022

Available from 08/17/2022

SamenKhan
SamenKhan 🇵🇰

231 documents

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bg1
Potential computations for
uniform charge distributions
•Potential of a spherical conductor
•E Field of a charged spherical conductor by G law, same as
for a point charge( E field is ‘0’ inside the sphere)
E = k q / r2
•Potential V = K q / R
•E field just at the outside surface
E = K q / R2 divide V by E
•V / E = K q R2 / K q R = R
•V = E R or E = V / R
R
E= kq/r2
V= kq/r
V= kq/R
E= kq/R2
pf3
pf4
pf5
pf8
pf9

Partial preview of the text

Download POTENTIAL POINT IN APPLIED PHYSICS and more Slides Physics in PDF only on Docsity!

Potential computations for

uniform charge distributions

  • (^) Potential of a spherical conductor
  • (^) E Field of a charged spherical conductor by G law, same as

for a point charge( E field is ‘0’ inside the sphere)

E = k q / r

2

  • (^) Potential V = K q / R
  • (^) E field just at the outside surface

E = K q / R

2

divide V by E

  • (^) V / E = K q R^2 / K q R = R
  • (^) V = E R or E = V / R R E= kq/r^2 V= kq/R V= kq/r E= kq/R^2
  • (^) When E = E max

= 0.8 x

6

NC

, air starts conducting, then

potential V = V

max

, i.e. Potential bearing capacity of a

conductor V = E R

  • (^) Terminals of electrical machines(transformer, batteries) are

circular

  • (^) Pointed terminals, E field is large even at lower potential

and air starts conducting (leakage) when E Fd > E

max

  • (^) Example: Terminal of an Electrical device is 0.5 cm, find

max potential it can bear

V

max

= E

max

R = 0.8 x

6

Nc

x 0.005 m = 4 KV

If radius R = 0.5m, then V

max

= 400 KV

Example

  • (^) Example: Terminals of a 6 V battery are separated 0.5 m apart. Find E field between the terminals.
  • (^) E = V ab / L^ in this case^ E = 6 J C - / 0.5 m =12 NC - (N m C - / m)
  • (^) Example: Find the velocity of electron in an electron gun or a diode when Cathode potential = 0 V, Anode potential = 250 V Electron is released from the electron gun vc = 0 ms

and accelerated towards anode Va = 250 V, then work done in terms of PD:

  • (^) W = e (Va -Vc) = = – (0)
  • (^) (1.6 x 10-19 C) (250 j C-1) =
  • (^) = [ 2(1.6 x 10-19 C) (250 j C-1) / (9.1 x10-31Kg)]1/
  • (^) we =9.4 x 106 m s-1^ ( N m/ Kg)1/2^ = Kg m s-2x m / kg )1/

Potential computations for

uniform charge distributions

  • (^) Potential outside the surface of a charged cylinder / line of charge We know by Gauss’s Law PD between any two points a & b at radial distance ra & rb If b is at then vb=0 and the eq becomes Hence ra= is not suitable. We can assume v=0 at some distance ro Then This gives values for r > R (radius of the cylinder) Then

GRADIENT OF POTENTIAL

  • (^) E field and potential in a region of space are always function of distance i.e. coordinates (x,y,z) of the point where being measured & related by Line integral
  • (^) The above equation can be modified to another useful form that relates components of E to infinitesimal variation of V i.e. dV derivatives of V
  • (^) represents small change in potential due to small displacement
  • (^) gives component of to When a charge (+ve) moves from higher to lower potential,W is – ve ab. Hence we can show it by putting a – ve sign before the integral
  • (^) We can rewrite line integral in terms of change in potential due to change in displacement
  • (^) The two integrals are equal for any pair of limits a and b, integrands must be equal, so
  • (^) or where is the rate of change of potential in the direction of
  • (^) If has a component to x axis, then component of is