How much work does the electric field do in moving a 2 μC charge from a point with
a potential of +20 V to a point with a potential of −20 V?
W=q(φ1-φ2) = 2•10^-6•(20-(-20))=8•10^-5 J.
To calculate the work done by the electric field in moving a charge from one point to another, we can use the formula:
Work = Charge × Potential Difference
In this case, the charge is given as 2 μC (microcoulombs), the potential at the initial point is +20 V, and the potential at the final point is -20 V.
First, let's convert the charge from microcoulombs to coulombs. 1 μC is equal to 10^-6 C, so 2 μC is equal to 2 × 10^-6 C.
Now we can calculate the potential difference:
Potential Difference = Final Potential - Initial Potential
= (-20 V) - (+20 V)
= -40 V
Note that the negative sign indicates a potential difference in a direction opposite to that of the electric field.
Finally, we can plug in the values into the formula:
Work = Charge × Potential Difference
= (2 × 10^-6 C) × (-40 V)
= -80 × 10^-6 J
= -8 × 10^-5 J
Therefore, the work done by the electric field in moving the 2 μC charge from a point with a potential of +20 V to a point with a potential of −20 V is -8 × 10^-5 J (joules).