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Transcript
Book Reference : Pages 86-88 1. To understand what we mean by “point charge” 2. To consider field strength as a vector 3. To apply our knowledge of equipotentials to electric fields 4. To understand what is meant by potential gradients We can consider a charge to be a “point charge” if.... Point charge Q & test charge q 1. 2. 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 +Q r +q This is comparable to assumptions made about the separation and diameter of planets during gravitation Coulomb’s law gives us the force : F = 1 Qq 40 r2 By definition the electric field strength (E= F/q) making the electric field strength at a distance r from Q E= 1 Q 40 r2 Note if Q is negative, this formula will yield negative numbers indicating that the field lines are pointing inwards 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 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: 1. Forces in the same direction : F1 -Q1 point charge F2 Test charge +q +Q2 point charge 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 2. Forces in the opposite direction : F1 +Q1 point charge F2 Test charge +q +Q2 point charge 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 = E 1 - E2 3. Forces at right angles: Test charge +q F1 F2 +Q1 point charge -Q2 point charge Standard resolving techniques... From Pythagoras F2 = F12 + F22 Electric Field Strength E2 = E12 + E22 Trigonometry can be used to find the resultant direction 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) +Q A test charge moving along an equipotential has constant potential energy & so no work is done by the electric field Equipotential lines and field lines always meet at right angles +Q Y X +600 V +1000 V +400V Ep = QV 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 2C is moved from X to Y 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 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 Uniform field : When the field is uniform, (e.g. Between oppositely -ve plate charged parallel plates) then the equipotentials are 0 equally spaced and parallel to the plates Equipotential Lines +ve plate +V Potential V 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) Distance d