A uniform electric field of magnitude 250 V/m is directed in the positive x-direction. A +12 μC charge moves from the origin to the point (x, y)=(20.0 cm,5.0 cm). What is the total work done required in moving the charge particle to the new position?

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Gerardo Goodman · Accepted Answer

Given,Electric field,   E  =  250     V      /    mCharge,   q  =  12     μ  C  =  12  ×      10          −      6           CSince the entire plane has the same potential regardless of how we move along the y-axis, there is no change in potential; however, as we move along the x-axis, there will be a change in potential, and this change in potential energy will be      V    2    −      V    1    =→  E  →  dSubstituting the values given in the question,  ⇒      V    2    −      V    1    =  −  250  ×  20  ×      10          −      2          ⇒      V    2    −      V    1    =  −  50     VSo, workdone in moving charge q, from origin to new position,  W  =  q  (      V    2    −      V    1    )    ⇒  W  =  12  ×      10          −      6        ×  (  −  50  )    ⇒  W  =  −  6  ×      10          −      4           J    ⇒  W  =  −  0.6     m  J

A uniform electric field of magnitude 250 V/m is directed in the positive x-direction. A +12 μC charge moves from the origin to the point (x, y)=(20.0 cm,5.0 cm). What is the total work done required in moving the charge particle to the new position?

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