How much work does the electric field do in moving a -7.7 micro Coulomb charge from ground to a point whose potential is +55 V higher?
To calculate the work done by the electric field in moving a charge from one point to another, we need to use the formula:
Work = q * ΔV
where:
- Work is the work done (in joules),
- q is the charge being moved (in coulombs),
- ΔV is the potential difference between the two points (in volts).
Given:
- q = -7.7 μC (microcoulombs)
- ΔV = +55 V
Let's plug in the values into the formula and calculate the work done:
Work = (-7.7 μC) * (+55 V)
= -7.7 μC * 55 V
To multiply these values, we need to convert the charge into coulombs by dividing it by 1,000,000, and then multiply it by 55:
Work = (-7.7 μC / 1,000,000) * 55 V
= -(-0.0000077 C) * 55 V
= 0.0000077 C * 55 V
= 0.0004235 C·V
Therefore, the work done by the electric field in moving a -7.7 μC charge from ground to a point whose potential is +55 V higher is approximately 0.0004235 C·V (coulomb-volts).
To calculate the work done by the electric field in moving a charge, you can use the formula:
Work = Charge * Voltage
In this case, the charge is -7.7 micro Coulombs and the potential difference (voltage) is +55 V.
First, convert the charge from micro Coulombs to Coulombs:
-7.7 micro Coulombs = -7.7 x 10^-6 Coulombs
Now, plug the values into the formula:
Work = (-7.7 x 10^-6 Coulombs) * (+55 V)
Calculating the multiplication:
Work = -4.235 x 10^-4 Coulomb-Volts
The unit Coulomb-Volts is also known as a Joule (J), which is the unit for work.
Therefore, the work done by the electric field in moving the -7.7 micro Coulomb charge from ground to a point with a potential +55 V higher is approximately -4.235 x 10^-4 Joules.