Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law
so voltage across the capacitor + voltage across resistor = V
Now we know that
here rate of change in energy of the capacitor is given as
In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 × )(1.2 × )
PE = 5.0 x joules
None of the options shown is correct.
Answer : D) Circuit A
This circuit is the only circuit where it is not complete, having and open spot towards the bottom of it, making it and open circuit.