PHY Electrostatics: PLEASEEE help! Determine the magnitude and direction of the resultant electric force on q
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PHY Electrostatics: PLEASEEE help! Determine the magnitude and direction of the resultant electric force on q

[From: ] [author: ] [Date: 12-07-09] [Hit: ]
536 * ke*q^2/a^2You arent supposed to plug in the values of ke, q, and a, because it tells you to leave them in symbolic form.Even though the value of ke is a universal constant.Be sure to indicate that both q and a are each squared.......
| Fnet | = ke*q/a^2 * sqrt((Qb + Qc*sqrt(2)/4)^2 + (Qb + Qc*sqrt(2)/4)^2 )

Simplify:
| Fnet | = ke*q*(Qb + Qc*sqrt(2)/4)*sqrt(2)/a^2

Data:
Qb = 2.5*q
Qc = 6*q


| Fnet | = ke*q^2*(2.5 + 6*sqrt(2)/4)*sqrt(2)/a^2


Gather:
| Fnet | = (ke*q^2/a^2) *(2.5 + 6*sqrt(2)/4)*sqrt(2)

Crunch the numeric part of the expression:
(2.5 + 6*sqrt(2)/4)*sqrt(2) = 6.53553391


Answer:
| Fnet | = 6.536 * ke*q^2/a^2

You aren't supposed to plug in the values of ke, q, and a, because it tells you to leave them in symbolic form. Even though the value of ke is a universal constant. Be sure to indicate that both q and a are each squared.

45 deg CCW of +x is of course your direction. We know this, because the vector terms are both positive and identical. We also know the direction, just by inspection, because of the symmetry of the problem.
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