Timeline for Why there is a negative sign in the formula of calculating work done in electrostatics?
Current License: CC BY-SA 4.0
14 events
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May 25, 2019 at 13:56 | answer | added | Bob D | timeline score: 0 | |
May 24, 2019 at 16:33 | comment | added | The Photon | it isn't the applied force it is the "electrical force". Please see my answer. | |
May 24, 2019 at 16:29 | answer | added | The Photon | timeline score: 3 | |
May 24, 2019 at 16:27 | answer | added | Shreyansh Pathak | timeline score: 0 | |
May 24, 2019 at 16:17 | history | edited | Hawkingo | CC BY-SA 4.0 |
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May 24, 2019 at 16:13 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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May 24, 2019 at 16:12 | history | reopened | Qmechanic♦ | ||
May 24, 2019 at 16:09 | comment | added | Hawkingo | Then if I change the coordinate, for example, the other charge is moving in q3 quadrant ( in 2 dimensions, (x and y are negative)) then the dr will be positive, so how can a formula change /depends on a coordinate? | |
May 24, 2019 at 16:06 | history | edited | Hawkingo | CC BY-SA 4.0 |
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May 24, 2019 at 16:01 | comment | added | Hawkingo | The $ \vec { F}$ here is the applied force | |
May 24, 2019 at 16:01 | history | closed | Qmechanic♦ | Needs details or clarity | |
May 24, 2019 at 16:00 | comment | added | The Photon | How is $\vec{F}$ defined in this problem? | |
May 24, 2019 at 15:30 | answer | added | PhysicsMan | timeline score: 0 | |
May 24, 2019 at 15:14 | history | asked | Hawkingo | CC BY-SA 4.0 |