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www.GOALias.blogspot.comPhysics2h2hmpt = p.a= eE= ×p–713 10 sThus, the heavier particle (proton) takes a greater time to fall throughthe same distance. This is in basic contrast to the situation of ‘freefall under gravity’ where the time of fall is independent of the mass ofthe body. Note that in this example we have ignored the accelerationdue to gravity in calculating the time of fall. To see if this is justified,let us calculate the acceleration of the proton in the given electricfield:apeE=m=p(1. 6 ×−19 10 C) × (2.0 ×410−1N C )−27167 . × 10 kg22EXAMPLE 1.9 EXAMPLE 1.812 –2= 19 . × 10 mswhich is enormous compared to the value of g (9.8 m s –2 ), theacceleration due to gravity. The acceleration of the electron is evengreater. Thus, the effect of acceleration due to gravity can be ignoredin this example.Example 1.9 Two point charges q 1and q 2, of magnitude +10 –8 C and–10 –8 C, respectively, are placed 0.1 m apart. Calculate the electricfields at points A, B and C shown in Fig. 1.14.FIGURE 1.14Solution The electric field vector E 1Aat A due to the positive chargeq 1points towards the right and has a magnitudeE9 2 -2 −8(9 × 10 Nm C ) × (10 C)1A 2= = 3.6 × 10 4 N C –1(0.05 m)The electric field vector E 2Aat A due to the negative charge q 2pointstowards the right and has the same magnitude. Hence the magnitudeof the total electric field E Aat A isE A= E 1A+ E 2A= 7.2 × 10 4 N C –1E Ais directed toward the right.