Answer is not 1117.93 m/s A 32.91-mC charge is placed 37.65 cm to the left of a

Question

Answer is not 1117.93 m/s

A 32.91-mC charge is placed 37.65 cm to the left of a 88.73-mC charge, as shown in the figure, and both charges are held stationary. A particle with a charge of -6.051 mu C and a mass of 27.31 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 v_A at that point?

Explanation / Answer

here,

q1 = 0.03291 C

q2 = 0.08873 C

q = - 6.051 * 10^-6 C

potential at initial point , Vi = k * q1*q/( sqrt(0.3765^2 + 0.3389^2)) + k * q2 * q /( 0.3389)

potential at A , Va = k * q1*q/(0.113) + k * q2 * q /( 0.2635)

let the speed at a be va

using work energy theorm

work done by potntial = kinetic energy

( Va - Vb) * q = 0.5 * m * va^2

(k * q1*q/(0.113) + k * q2 * q /( 0.2635) - k * q1*q/( sqrt(0.3765^2 + 0.3389^2)) + k * q2 * q /( 0.3389) ) * q = 0.5 * 0.02731 * va^2

9 * 10^9 *(6.051 * 10^-6)^2 * (0.03291*/(0.113) + 0.08873 /( 0.2635) - 0.032291/( sqrt(0.3765^2 + 0.3389^2)) + 0.08873 /( 0.3389) ) = 0.5 * 0.02731 * va^2

va = 4.46 m/s

the speed at a is 4.46 m/s

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