If the path of the particle were to pass through the gray point labeled A, what

Question

If the path of the particle were to pass through the gray point labeled A, what would be its speed at that point?

bod's chames ae id plaionas7.so patide welha char3e 7:s tareeanas ashnan nf 12 8rge, and A 33.91-mC charge is placed 37.65 cm to the left of a 73.73-mC charge, as shown in the figure, and both charges are held stationary. A particle with a charge of-3.551 C and a mass of 42.81 g (depicted as a blue sphere) is placed at rest at a distance 33.89 cm above the right-most charge. If the particle were to be released from rest, it would follow some complicated path around the two stationary charges Calculating the exact path of the particle would be a challenging problem, but even without performing such a calculation it is possible to make some definite predictions about the future motion of the particle. If the path of the particle were to pass through the gray point labeled A what would be its speed VA at that point? Number 3.551 pC m/s 33.89 cm +33.91 mC k-11.30cm K37.65 cm +73.73 mC

Explanation / Answer

Using conservation of energy

At initial position particle is at rest hence will have just potential energy and zero kinetic energy

At A it will have both

Hence

KEi +PEi =KEf +PEf

Or change

PEi - PEf = KEf

PEi

qVi

Now distance between the charge on left and blue charge can be calculated using pythagorus theorem

= 50.66cm

Calculating Vi =

k*73.73*10-3/0.3389 + k*33.91*10-3/0.5066 = k*0.2845 V

PEi = -3.551*10-6*k*0.2845 J

Now for PEf = qVf

Distance between right charge at blue charge in final position = 37.65-11.30=26.35cm

Vf = k*73.73*10-3/0.2635 + k*33.91*10-3/0.1130 = k*0.580 V

PEf = -3.551*10-6*k*0.580 J

Using

PEi - PEf = KEf

-3.551*10-6*k*0.2845 - -3.551*10-6*k*0.580 = KEf

9440.65 =KEf

0.5*0.04281v2 = 9440.65

v = 664.1m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---