Electric Potential Energy and Electric Potential Difference

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To calculate the work required to move a charge of -4.0 microcoulombs between charged parallel plates connected to a 12-V battery, the formula used is Work (w) = charge (q) x voltage (V), resulting in w = -4.8 x 10^-5 J. The negative sign indicates that the work is done against the repulsive force. However, additional information such as the capacitance of the plates, their area, and the distance between them is necessary for a complete analysis. The electric potential created by the battery may not lead to a uniform electric field, which complicates the calculation. Accurate work determination requires more specific details about the system.
predentalgirl1
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A pair of parallel plates is charged by a 12-V battery. How much work is required to move a particle with a charge of --4.0 microcoulombs (MC) from the positive to the negative plate?








Given that,
Emf = 12 V
q = -4 x 10 -6 C
I have,
Work done (w) = q x V
= -4 x 10 -6 x 12
= - 4.8 x 10-5 J ?
‘-‘sign indicates that work is done against the force of
repulsion

Is my work/answer correct??
 
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help please!
 
you seem to be missing some information. what is the capacitance of the parallel plate? or, what is the area of the plates and the distance between the plates?

YOU CAN'T ASSUME THAT THE ELECTRIC POTENTIAL OF THE BATTERY IS GOING TO CREATE A PERFECT CHARGE DISTRIBUTION TO CREATE AN ELECTRIC FIELD NEEDED TO FIGURE OUT THE WORK. IN OTHER WORDS...

V DOES NOT EQUAL -WORK/Q WITH THE INFORMATION YOU'VE GIVEN UNLESS YOU'RE TAKING SOME TYPE OF SURVEY COURSE...HAHAHA.
 
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