A charge of #8 C# is passing through points A and B on a circuit. If the charge's electric potential changes from #32 J# to #4 J#, what is the voltage between points A and B?

Question

A charge of #8 C# is passing through points A and B on a circuit.  If the charge's electric potential changes from #32 J# to #4 J#, what is the voltage between points A and B?

Adrian Ayala · Accepted Answer

The voltage is #=3.5V# The potential difference between the points #A# and #B# is #=U# Charge/Energy = Voltage #U=(DeltaE)/Q# #DeltaE=32-4=28J# The charge is #Q=8C# #1 V=(1J)/C# #U=28/8=3.5V#

Xavier Adame · Answer

undefined The voltage between points A and B can be calculated using the formula:

V = ΔE / q

Where:
- V is the voltage
- ΔE is the change in electric potential energy (final electric potential energy minus initial electric potential energy)
- q is the charge

Given that the charge q = 8 C, and the change in electric potential energy ΔE = 32 J - 4 J = 28 J, we can plug these values into the formula:

V = 28 J / 8 C

V ≈ 3.5 V

So, the voltage between points A and B is approximately 3.5 volts.

Axel Acevedo · Answer

undefined The voltage between points A and B can be calculated using the formula:

$$ V = \frac{W}{Q} $$

where $ V $ is the voltage, $ W $ is the work done on the charge, and $ Q $ is the charge.

Given that the electric potential changes from 32 J to 4 J, the work done on the charge ($ W $) is the difference in electric potential, which is $ 32 J - 4 J = 28 J $.

So, $ W = 28 $ J and $ Q = 8 $ C.

Using the formula:

$$ V = \frac{W}{Q} = \frac{28 \, J}{8 \, C} = 3.5 \, V $$

Therefore, the voltage between points A and B is 3.5 volts.