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Point Charges

Learning Objectives. Book Reference : Pages 86-88. Point Charges. To understand what we mean by “point charge” To consider field strength as a vector To apply our knowledge of equipotentials to electric fields To understand what is meant by potential gradients.

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Point Charges

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  1. Learning Objectives Book Reference : Pages 86-88 Point Charges To understand what we mean by “point charge” To consider field strength as a vector To apply our knowledge of equipotentials to electric fields To understand what is meant by potential gradients

  2. We can consider a charge to be a “point charge” if.... Point Charges 1 Point charge Q & test charge q • The separation of the objects is much greater than the size of the object • If its charge does not affect the electric field it is in • This is comparable to assumptions made about the separation and diameter of planets during gravitation +Q r +q

  3. Coulomb’s law gives us the force : • F = 1 Qq • 40 r2 • By definition the electric field strength (E= F/q) making the electric field strength at a distance r from Q • E = 1 Q • 40 r2 Point Charges 2 Note if Q is negative, this formula will yield negative numbers indicating that the field lines are pointing inwards

  4. Calculate the electric field strength 0.35nm away from a nucleus with a charge of +82e 0 = 8.85 x 10-12 F/m e = 1.6 x 10-19 C Worked Example

  5. If our test charge is in an electric field due to multiple charges each exerts a force. The resultant force per unit charge (F/q) gives the resultant field strength at the particular position of our test charge We can consider 3 scenarios: Field Strength as a Vector 1

  6. 1. Forces in the same direction : Our test charge experiences two forces F1 = qE1 & F2 = qE2 The resultant F is simply F = F1 + F2 The resultant field strength E = F/q = (qE1 + qE2) /q E = E1 + E2 Field Strength as a Vector 2 F1 F2 -Q1 point charge Test charge +q +Q2 point charge

  7. 2. Forces in the opposite direction : Our test charge experiences two forces F1 = qE1 & F2 = qE2 The resultant F is simply F = F1 - F2 The resultant field strength E = F/q = (qE1 - qE2) /q E = E1 - E2 Field Strength as a Vector 3 F1 F2 +Q1 point charge Test charge +q +Q2 point charge

  8. 3. Forces at right angles: Standard resolving techniques... From Pythagoras F2 = F12 + F22 Electric Field Strength E2 = E12 + E22 Trigonometry can be used to find the resultant direction Field Strength as a Vector 4 F1 Test charge +q F2 +Q1 point charge -Q2 point charge

  9. Equipotentials are lines of constant potential & can be compared to contour lines on a map. (and are the same as we have encountered for gravitation) Equipotentials 1 A test charge moving along an equipotential has constant potential energy & so no work is done by the electric field +Q Equipotential lines and field lines always meet at right angles

  10. Note the lines of equal potential (measured in V) are shown by the equipotential lines Consider the change in potential energy if a test charge of 2C is moved from X to Y Equipotentials 2 Y X +600 V +1000 V +Q +400V Ep= QV at 1000V Ep = 2x10-6 x 1000 = 2x10-3J at 400V Ep = 2x10-6 x 400 = 8x10-4J The change in potential energy is 1.2x10-3J

  11. Definition : The potential gradient is the change in potential per unit change in distance in a given direction Two scenarios: Non uniform field : The potential gradient varies according to position & direction. The closer the equipotentials the greater the potential gradient Potential Gradient 1

  12. Uniform field : When the field is uniform, (e.g. Between oppositely charged parallel plates) then the equipotentials are equally spaced and parallel to the plates Graph shows that potential relative to the –ve plate is proportional to distance (pg is constant & is V/d) (Potential increases opposite direction to field) Equipotential Lines Potential Gradient 2 -ve plate +ve plate 0 +V Potential V Distance d

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